🧠:[<think>Okay, so I need to find the minimum distance from a point P on the curve y = x² - ln x to the straight line y = x - 2. Hmm, let me think about how to approach this. First, I remember that the distance from a point (x, y) to a line ax + by + c = 0 is given by the formula |ax + by + c| / sqrt(a² + b²). So maybe I should start by rewriting the given line equation into the standard form. The line is y = x - 2. Let me rearrange that. Subtract y and add 2 to both sides: x - y - 2 = 0. So the line is x - y - 2 = 0. Therefore, a = 1, b = -1, c = -2. So, the distance from a point (x, y) on the curve to this line would be |x - y - 2| / sqrt(1 + (-1)²) = |x - y - 2| / sqrt(2). But since the point P is on the curve y = x² - ln x, I can substitute y with x² - ln x in the distance formula. So substituting y, the distance becomes |x - (x² - ln x) - 2| / sqrt(2) = |x - x² + ln x - 2| / sqrt(2). Since we want the minimum distance, we can ignore the sqrt(2) in the denominator for now because minimizing the numerator will also minimize the entire expression. So essentially, we need to minimize the function D(x) = |x - x² + ln x - 2|. But dealing with absolute values can be tricky when taking derivatives. However, since the square function is monotonic for non-negative values, we can instead minimize the square of the distance to make differentiation easier. Wait, but actually, since the absolute value complicates things, maybe we can consider the expression inside the absolute value and find where it's minimized. However, depending on the sign of the expression, the absolute value could flip the sign, so perhaps it's safer to square the distance function to remove the absolute value. Alternatively, since we are looking for the minimal distance, which is a non-negative quantity, maybe the minimal distance occurs where the expression inside the absolute value is either minimized or maximized depending on the sign. But perhaps squaring the distance is the standard approach here.But actually, the minimal distance will occur when the derivative of the distance function is zero. Since the square root and the absolute value are both involved, taking the derivative directly might be complex. Let me think again.Alternatively, since distance is a non-negative function, the minimal distance will occur at the same x-value where the square of the distance is minimal. So maybe it's easier to square the distance function and then minimize that. Let's try that.So squared distance S(x) = [ (x - x² + ln x - 2)² ] / 2. To minimize S(x), we can take its derivative, set it to zero, and solve for x. Wait, but actually, since we divided by sqrt(2), squaring the distance would give S(x) = [ (x - x² + ln x - 2)² ] / 2. But to make things simpler, since 1/sqrt(2) is a constant multiplier, minimizing S(x) is equivalent to minimizing (x - x² + ln x - 2)². So we can focus on minimizing the numerator squared.But even better, since squaring is a monotonic transformation for non-negative values, minimizing the squared distance is equivalent to minimizing the distance itself. So let's proceed by squaring the expression inside the absolute value, but since the absolute value is squared, that's the same as squaring the original expression. So S(x) = (x - x² + ln x - 2)^2. But actually, wait. The original distance is |x - y - 2| / sqrt(2), and since y = x² - ln x, the expression inside the absolute value is x - (x² - ln x) - 2 = x - x² + ln x - 2. So S(x) = (x - x² + ln x - 2)^2. To find the minimum, take the derivative of S(x) with respect to x, set it to zero, and solve.Let me compute S'(x). Let's denote f(x) = x - x² + ln x - 2, so S(x) = [f(x)]². Then S'(x) = 2 f(x) f'(x). First, compute f'(x): derivative of x is 1, derivative of -x² is -2x, derivative of ln x is 1/x, derivative of -2 is 0. So f'(x) = 1 - 2x + 1/x.So S'(x) = 2 (x - x² + ln x - 2)(1 - 2x + 1/x). To find critical points, set S'(x) = 0. Therefore, either f(x) = 0 or f'(x) = 0.But f(x) = 0 would mean that the distance is zero, i.e., the point P is on the line y = x - 2. Let's check if such a point exists on the curve. So solving x - x² + ln x - 2 = 0. That would be x² - ln x = x - 2, so y = x - 2. But is there a solution to x² - ln x = x - 2? Let's check for x > 0 (since ln x is defined only for x > 0).Let me test x = 1: Left side: 1 - 0 = 1; right side: 1 - 2 = -1. Not equal.x = 2: Left: 4 - ln 2 ≈ 4 - 0.693 ≈ 3.307; right: 2 - 2 = 0. Not equal.x = 3: Left: 9 - ln 3 ≈ 9 - 1.098 ≈ 7.902; right: 3 - 2 = 1. Not equal.x approaching 0+: ln x approaches -infinity, so left side x² - ln x approaches infinity. Right side x - 2 approaches -2. So left side is much larger. x approaching infinity: x² dominates, so left side is x² - ln x ~ x²; right side is x - 2 ~ x. So left side is much larger. Therefore, it seems that f(x) = x - x² + ln x - 2 is always negative? Let's see. Let's compute f(x) at x = 1: 1 - 1 + 0 - 2 = -2. At x = 2: 2 - 4 + ln 2 - 2 ≈ -4 + 0.693 ≈ -3.307. At x = 0.5: 0.5 - 0.25 + ln 0.5 - 2 ≈ 0.25 - 0.693 - 2 ≈ -2.443. So yes, f(x) is negative for all x > 0, so |f(x)| = -f(x). Therefore, the distance is |f(x)| / sqrt(2) = (-f(x)) / sqrt(2). Therefore, since f(x) is negative everywhere, we can instead minimize -f(x) = x² - x - ln x + 2. Wait, but actually, since we want to minimize the distance, which is proportional to |f(x)|, but since f(x) is negative, |f(x)| = -f(x). Therefore, minimizing |f(x)| is equivalent to minimizing -f(x) = x² - x + ln x - 2. Wait, hold on:Wait, f(x) = x - x² + ln x - 2, so -f(x) = x² - x - ln x + 2. Therefore, the distance is |f(x)| / sqrt(2) = (x² - x - ln x + 2)/sqrt(2). So to minimize the distance, we can minimize the numerator x² - x - ln x + 2. Let me verify this.Alternatively, perhaps I should just work with the original expression. Since f(x) is negative, the absolute value makes it positive, so the distance is (-f(x))/sqrt(2) = (x² - x - ln x + 2)/sqrt(2). Therefore, to minimize the distance, we can instead minimize the function g(x) = x² - x - ln x + 2. Alternatively, perhaps not, because if f(x) is negative, then |f(x)| = -f(x), so the distance is (-f(x))/sqrt(2) = (x² - x - ln x + 2)/sqrt(2). Therefore, to minimize the distance, we need to minimize x² - x - ln x + 2. But maybe instead of dealing with the absolute value, it's simpler to note that since the distance is |linear combination|, the minimal distance occurs when the derivative of the square of the distance is zero, regardless of the sign. So perhaps we can proceed with minimizing [f(x)]², which would take into account both the positive and negative regions, but since f(x) is always negative, squaring it would just be the square of a negative number, but the minimal square would occur at the minimal |f(x)|. So perhaps, even though f(x) is always negative, squaring it and then minimizing will still find the minimal |f(x)|. Therefore, maybe I should just proceed with minimizing [f(x)]². So let's stick with S(x) = (x - x² + ln x - 2)^2. Then take the derivative S'(x) = 2*(x - x² + ln x - 2)*(1 - 2x + 1/x). Setting this equal to zero. Therefore, either the first factor is zero, which we saw leads to no solution, or the second factor is zero. So, since f(x) is never zero, as we saw, the critical points must come from setting the second factor to zero: 1 - 2x + 1/x = 0. So, 1 - 2x + 1/x = 0. Let's solve this equation for x > 0. Multiply both sides by x to eliminate the denominator: x*(1) - 2x² + 1 = 0. So, x - 2x² + 1 = 0. Rearranged: -2x² + x + 1 = 0. Multiply both sides by -1: 2x² - x - 1 = 0. Now, solve the quadratic equation 2x² - x - 1 = 0. Using the quadratic formula: x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3] / 4. Therefore, x = (1 + 3)/4 = 1 or x = (1 - 3)/4 = -0.5. But since x > 0, we discard x = -0.5. Therefore, the critical point is at x = 1. So, x = 1 is the critical point. Now, we need to check if this is a minimum. To confirm, we can use the second derivative test or check the sign changes of the first derivative. Let me compute the second derivative. First, S'(x) = 2*(f(x))*(f’(x)). So S''(x) would be 2*[f’(x)*f’(x) + f(x)*f''(x)] by the product rule. But this might get complicated. Alternatively, since we have only one critical point (x=1), we can check the value of S(x) around x=1. Let me compute S(x) at x=1 and at points slightly less than and greater than 1. First, at x = 1: f(1) = 1 - 1 + ln 1 - 2 = 0 + 0 - 2 = -2. So S(1) = (-2)^2 = 4. Now, let's try x = 0.5: f(0.5) = 0.5 - 0.25 + ln 0.5 - 2 ≈ 0.25 - 0.693 - 2 ≈ -2.443. S(0.5) ≈ (-2.443)^2 ≈ 5.967. At x = 2: f(2) = 2 - 4 + ln 2 - 2 ≈ -4 + 0.693 ≈ -3.307. S(2) ≈ (-3.307)^2 ≈ 10.938. At x = 0.9: f(0.9) = 0.9 - 0.81 + ln 0.9 - 2 ≈ 0.09 - 0.105 - 2 ≈ -2.015. S(0.9) ≈ (-2.015)^2 ≈ 4.06. At x = 1.1: f(1.1) = 1.1 - 1.21 + ln 1.1 - 2 ≈ -0.11 + 0.095 - 2 ≈ -2.015. S(1.1) ≈ (-2.015)^2 ≈ 4.06. Wait, so at x = 0.9 and x = 1.1, S(x) is approximately 4.06, which is slightly higher than S(1) = 4. So this suggests that x=1 is indeed a local minimum. Since it's the only critical point and the function S(x) tends to infinity as x approaches 0 or infinity, x=1 is the global minimum. Therefore, the minimal distance occurs at x=1. Now, compute the minimal distance. The distance is |f(x)| / sqrt(2) = | -2 | / sqrt(2) = 2 / sqrt(2) = sqrt(2). So the minimal distance is sqrt(2). But let me verify this again. At x=1, the point P on the curve is (1, y). y = (1)^2 - ln 1 = 1 - 0 = 1. So point P is (1,1). The distance from (1,1) to the line y = x - 2. Using the distance formula: |1 - 1 - 2| / sqrt(1 + 1) = | -2 | / sqrt(2) = 2 / sqrt(2) = sqrt(2). Yes, that checks out. But let me double-check my steps. When I found the critical point at x=1, that gave the point (1,1). The distance from (1,1) to the line y = x - 2 is indeed sqrt(2). Wait, but just to make sure there's no mistake in solving 1 - 2x + 1/x = 0. Let me verify that solution again. Starting with 1 - 2x + 1/x = 0. Multiply by x: x - 2x² + 1 = 0. So -2x² + x + 1 = 0. Multiply by -1: 2x² - x - 1 = 0. Solutions x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4, which gives x=1 and x=-0.5. Correct, so x=1 is the only valid solution. Therefore, the minimal distance is sqrt(2). But wait, let me check another point for sanity check. Suppose I take x=1. Let's see if moving slightly away from x=1 increases the distance. For example, take x=1. Let’s compute the distance. At x=1, the point is (1,1). The line is y = x - 2. At x=1, the line has y = -1. So the vertical distance between (1,1) and (1,-1) is 2, but the actual minimal distance is not vertical. Wait, but according to our calculation, the distance is sqrt(2). Let me compute the distance using coordinates. The distance from (1,1) to the line y = x - 2. Using the formula: |x - y - 2| / sqrt(2) = |1 - 1 - 2| / sqrt(2) = |-2| / sqrt(2) = 2/sqrt(2) = sqrt(2). Correct. Alternatively, the distance from (1,1) to the line can be found by finding the perpendicular distance. Alternatively, parametrize the line y = x - 2. A point on the line can be written as (t, t - 2). The distance squared between (1,1) and (t, t - 2) is (t - 1)^2 + (t - 2 - 1)^2 = (t -1)^2 + (t -3)^2. To minimize this, take derivative with respect to t: 2(t -1) + 2(t -3) = 2t - 2 + 2t -6 = 4t -8. Set to zero: 4t -8 =0 → t=2. So the closest point on the line is (2, 0). The distance between (1,1) and (2,0) is sqrt((2-1)^2 + (0 -1)^2) = sqrt(1 +1) = sqrt(2). Correct. So that matches. Therefore, the minimal distance is indeed sqrt(2). Thus, after going through all these steps, I conclude that the minimal distance is sqrt(2).Final AnswerThe minimum distance from point P to the line ( y = x - 2 ) is (boxed{sqrt{2}}).</think>To find the minimum distance from a point ( P ) on the curve ( y = x^2 - ln x ) to the straight line ( y = x - 2 ), we start by expressing the distance from a point ((x, y)) to the line ( x - y - 2 = 0 ). The distance formula is given by:[text{Distance} = frac{|x - y - 2|}{sqrt{2}}]Since ( y = x^2 - ln x ), we substitute ( y ) into the distance formula:[text{Distance} = frac{|x - (x^2 - ln x) - 2|}{sqrt{2}} = frac{|x - x^2 + ln x - 2|}{sqrt{2}}]To minimize this distance, we need to minimize the expression inside the absolute value. Let ( f(x) = x - x^2 + ln x - 2 ). We find the critical points by taking the derivative of ( f(x) ) and setting it to zero:[f'(x) = 1 - 2x + frac{1}{x}]Setting ( f'(x) = 0 ):[1 - 2x + frac{1}{x} = 0]Multiplying through by ( x ):[x - 2x^2 + 1 = 0 implies 2x^2 - x - 1 = 0]Solving this quadratic equation using the quadratic formula:[x = frac{1 pm sqrt{1 + 8}}{4} = frac{1 pm 3}{4}]This gives the solutions ( x = 1 ) and ( x = -0.5 ). Since ( x > 0 ), we discard ( x = -0.5 ), leaving ( x = 1 ).We verify that ( x = 1 ) is a minimum by checking the second derivative or evaluating the function around ( x = 1 ). The point ( P ) on the curve at ( x = 1 ) is ( (1, 1) ). The distance from this point to the line ( y = x - 2 ) is:[frac{|1 - 1 - 2|}{sqrt{2}} = frac{|-2|}{sqrt{2}} = frac{2}{sqrt{2}} = sqrt{2}]Thus, the minimum distance from point ( P ) to the line ( y = x - 2 ) is (boxed{sqrt{2}}).,<think>Alright, so I need to find the minimum distance from a point P on the curve y = x² - ln x to the straight line y = x - 2. Hmm, okay. Let me start by recalling how to calculate the distance from a point to a line. The formula for the distance from a point (a, b) to the line Ax + By + C = 0 is |Ax + By + C| / sqrt(A² + B²). Right?First, I should probably write the equation of the given line in the standard form. The line is y = x - 2. If I rearrange this, it becomes x - y - 2 = 0. So here, A = 1, B = -1, and C = -2. Got that.Now, any point P on the curve y = x² - ln x will have coordinates (x, x² - ln x). Let's call this point (x, y), where y is defined by the curve. So, substituting into the distance formula, the distance D from P to the line is |1*x + (-1)*(x² - ln x) - 2| / sqrt(1² + (-1)²). Let me check that: substituting x and y into Ax + By + C. Since the line is x - y - 2 = 0, plugging in (x, y) gives x - (x² - ln x) - 2. The absolute value of that divided by sqrt(2). So yes, D = |x - (x² - ln x) - 2| / sqrt(2).But since we're looking for the minimum distance, we can focus on minimizing the numerator, because the denominator sqrt(2) is a positive constant and doesn't affect where the minimum occurs. So essentially, we can minimize the function f(x) = |x - x² + ln x - 2|. However, dealing with absolute values can complicate things. Maybe we can instead minimize the square of the distance, which would also have its minimum at the same point. Alternatively, since the absolute value just ensures positivity, we can consider the expression inside the absolute value and determine if it's always positive or negative. If it's always one sign, we can drop the absolute value and proceed.Let me check the expression inside the absolute value: x - x² + ln x - 2. Let's see for x > 0 (since ln x is only defined for x > 0). What does this function look like? Let's denote g(x) = x - x² + ln x - 2. We can analyze g(x) to see if it's always positive or negative, or changes sign.But maybe instead of worrying about the absolute value, since we need to minimize |g(x)|, perhaps we can square the distance function. The square of the distance would be [g(x)]² / 2. Then, minimizing [g(x)]² would also minimize the distance. Squaring might make differentiation easier, avoiding dealing with the absolute value.But either way, whether we minimize |g(x)| or [g(x)]², the critical points should be the same, except possibly when g(x) = 0. But since we're looking for a minimum, maybe the derivative approach can handle that.Alternatively, since calculus can handle absolute values by considering the derivative when the expression inside is positive or negative, but that might complicate things. Let's try squaring the distance function.So, squared distance S(x) = [ (x - x² + ln x - 2)^2 ] / 2. To minimize S(x), take derivative S'(x), set to zero.But before that, perhaps even better to just work with the original distance function and take its derivative. Let's see.But let me think again. Since the square root and the absolute value complicate things, perhaps squaring is better. Because squared distance is differentiable without dealing with square roots and absolute values. So squared distance is S(x) = (x - (x² - ln x) - 2)^2 / 2. Wait, no, the original distance is |g(x)| / sqrt(2), so squared distance is [g(x)]² / 2. Therefore, to minimize the squared distance, we can ignore the division by 2 and just minimize [g(x)]². So let's define S(x) = [g(x)]² = (x - x² + ln x - 2)^2. Then take derivative S'(x), set to zero.Alternatively, since S(x) is [g(x)]², S'(x) = 2 g(x) g'(x). Therefore, setting S'(x) = 0 implies either g(x) = 0 or g'(x) = 0. So critical points are either where the original function is zero (distance zero, which would be intersection points) or where the derivative of g(x) is zero. However, since we are dealing with a curve and a line, if they don't intersect, the minimal distance would be where the derivative of g(x) is zero.But let's check first if the curve and the line intersect. If they do, then the minimal distance would be zero at those points. So let's solve y = x² - ln x and y = x - 2. Setting x² - ln x = x - 2. So x² - x + 2 - ln x = 0. Let's see if this equation has any solutions for x > 0.Let me consider the function h(x) = x² - x + 2 - ln x. Find if h(x) = 0 for any x > 0. Let's analyze h(1): h(1) = 1 - 1 + 2 - 0 = 2. h(1) = 2 > 0. What about h approaching 0 from the right? As x approaches 0+, ln x approaches -infty, so -ln x approaches +infty, and x² - x approaches 0 - 0 = 0. Therefore, h(x) approaches +infty. At x = 1, h(x) is 2. What about x approaching +infty? Then x² dominates, so h(x) approaches +infty. So h(x) tends to +infty as x approaches 0+ and +infty. At x = 1, h(x) = 2. Maybe there's a minimum somewhere. Let's check the derivative of h(x): h'(x) = 2x - 1 - 1/x. Set derivative to zero: 2x - 1 - 1/x = 0. Multiply both sides by x: 2x² - x - 1 = 0. Solve quadratic: x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4. So x = 1 or x = -0.5. Since x > 0, x = 1 is critical point. So h(x) has a critical point at x = 1. But h(1) = 2, as computed earlier. So h(x) has a minimum at x = 1 of 2, which is positive. Therefore, h(x) is always positive for x > 0, meaning the curve y = x² - ln x and the line y = x - 2 do not intersect. Thus, the minimal distance is not zero, so we need to minimize |g(x)|, which is equivalent to minimizing [g(x)]².Since h(x) = x² - x + 2 - ln x is always positive, then g(x) = x - x² + ln x - 2 = - (x² - x + 2 - ln x) = -h(x). Therefore, g(x) is always negative, so |g(x)| = -g(x) = h(x). So the distance D(x) = h(x)/sqrt(2). Therefore, to minimize D(x), we need to minimize h(x). Since h(x) is the same as the function we already considered, which has a critical point at x = 1, and h(1) = 2, but we need to confirm whether that is a minimum.Wait, hold on. Wait, in the previous analysis, we found that h(x) has a critical point at x = 1, and since h''(x) can be calculated to check concavity. Let's compute h''(x). h'(x) = 2x -1 -1/x. Then h''(x) = 2 + 1/x². Since h''(x) is always positive for x > 0, the function h(x) is convex everywhere. Therefore, the critical point at x = 1 is a global minimum. Thus, the minimal value of h(x) is 2, achieved at x = 1. Therefore, the minimal distance is 2 / sqrt(2) = sqrt(2). But wait, that seems too straightforward. Wait, but if h(x) is minimized at x = 1 with value 2, then the minimal distance is 2 / sqrt(2) = sqrt(2). But let me verify this.Wait, but hold on. The distance D(x) is |g(x)| / sqrt(2) = | - h(x) | / sqrt(2) = h(x) / sqrt(2). Therefore, since h(x) is minimized at x =1 with h(1) =2, the minimal distance is 2 / sqrt(2) = sqrt(2). So the minimal distance is sqrt(2). But is this correct? Let me check.But wait, perhaps I made a confusion between h(x) and g(x). Let me go back. Original distance is |g(x)| / sqrt(2), where g(x) = x - x² + ln x -2. But h(x) was defined as x² - x + 2 - ln x, which is equal to -g(x). So h(x) = -g(x). Since we found that h(x) >=2 for all x>0, then g(x) = -h(x) <= -2. Therefore, |g(x)| = h(x). So D(x) = h(x)/sqrt(2). So since h(x) has a minimum of 2 at x=1, then minimal distance is 2 / sqrt(2) = sqrt(2). So answer is sqrt(2). But is this correct?Wait a second, but let's verify by plugging x =1 into the distance formula. At x=1, the point P is (1, 1² - ln 1) = (1, 1 - 0) = (1,1). The distance from (1,1) to the line y = x -2. Let's compute the distance. Using the formula |1 - 1 -2| / sqrt(1 +1) = |-2| / sqrt(2) = 2 / sqrt(2) = sqrt(2). So yes, that checks out.But wait, this seems too straightforward, but maybe that's the case. Wait, but I have to make sure that x=1 is indeed the point where the distance is minimized. Since h(x) is convex and has a unique minimum at x=1, then yes, that's the minimal value. Therefore, the minimal distance is sqrt(2). So is that the answer?But let me try to approach this problem again, perhaps using calculus on the distance function, just to confirm.Let me denote the distance from a point (x, y) on the curve to the line y = x -2. The distance is |x - y -2| / sqrt(2). Since y = x² - ln x, substitute that in: |x - (x² - ln x) -2| / sqrt(2) = | -x² + x + ln x -2 | / sqrt(2). Which is the same as |g(x)| / sqrt(2), where g(x) = -x² + x + ln x -2. So as before.To find the minimum of |g(x)|, since we already saw that g(x) is always negative (since h(x) = -g(x) >=2), then |g(x)| = -g(x) = x² -x - ln x +2. Wait, hold on, that contradicts? Wait, no:Wait, g(x) = -x² +x + ln x -2. Then, |g(x)| = | -x² +x + ln x -2 |. But h(x) = x² -x +2 - ln x = -g(x). Therefore, h(x) = -g(x), so |g(x)| = h(x). Therefore, to minimize |g(x)|, we need to minimize h(x). As before, h(x) has minimum at x=1, value 2, so |g(x)| has minimum 2, hence distance is 2 / sqrt(2) = sqrt(2). Therefore, the answer is sqrt(2). So that seems correct.But just to ensure, let's consider using calculus on the distance function directly, without substitution. Let's consider D(x) = |g(x)| / sqrt(2). Since we found that g(x) is always negative, D(x) = (-g(x)) / sqrt(2) = (x² -x - ln x +2)/sqrt(2). So we can consider f(x) = (x² -x - ln x +2). To minimize f(x), take derivative f'(x) = 2x -1 -1/x. Set to zero: 2x -1 -1/x =0. Multiply by x: 2x² -x -1=0. Which is the same quadratic as before. Solutions x=(1 ± sqrt(1 +8))/4=(1 ±3)/4. So x=1 or x=-0.5. Disregard negative x, so x=1. Then check second derivative: f''(x)=2 +1/x² >0, so convex. Therefore, x=1 is the minimum. Then f(1)=1 -1 -0 +2=2. Therefore, minimal distance is 2 / sqrt(2)=sqrt(2). Same result. So that seems consistent.Therefore, the minimal distance is sqrt(2). So the answer is sqrt(2). However, just to be thorough, let's check the behavior of the distance function at some other points. For example, let's pick x approaching 0+. Then y = x² - ln x approaches infinity, because ln x approaches -infty, so -ln x approaches +infty. Therefore, the point P is going far up, so the distance to the line y =x -2, which is a straight line with slope 1, would also approach infinity. Similarly, as x approaches infinity, y = x² - ln x ~x², so the point goes to infinity, and the distance to the line y =x -2 also approaches infinity. Therefore, the function D(x) must have a unique minimum at x=1. Therefore, the minimal distance is indeed sqrt(2).But let me also check a point near x=1. Let's say x=2. Compute the distance. Point P is (2, 4 - ln 2). Compute the distance to the line y =x -2. Using the formula |2 - (4 - ln2) -2| / sqrt(2) = |2 -4 + ln2 -2| / sqrt(2) = |-4 + ln2| / sqrt(2) ≈ |-4 +0.693| /1.414≈ |-3.307| /1.414≈3.307/1.414≈2.34, which is larger than sqrt(2)≈1.414. So yes, larger. Another point, x=0.5. Point P is (0.5, 0.25 - ln0.5). ln0.5≈-0.693. So y≈0.25 +0.693≈0.943. The distance is |0.5 -0.943 -2| / sqrt(2)= |-2.443| /1.414≈2.443/1.414≈1.727, which is still bigger than sqrt(2). So x=1 gives the minimal distance.Therefore, I think the answer is sqrt(2). But let me check once more.Wait, but when I plug x=1 into the distance formula, the point is (1,1), and the distance is |1 -1 -2| / sqrt(2) = |-2| / sqrt(2)=2/sqrt(2)=sqrt(2). Correct. So yes, the minimal distance is sqrt(2). Therefore, the answer is sqrt(2).Final AnswerThe minimum distance from point P to the line is boxed{sqrt{2}}.</think>To find the minimum distance from a point ( P ) on the curve ( y = x^2 - ln x ) to the straight line ( y = x - 2 ), we start by using the distance formula from a point to a line. The line ( y = x - 2 ) can be rewritten in standard form as ( x - y - 2 = 0 ). For a point ( (x, x^2 - ln x) ) on the curve, the distance ( D ) to the line is given by:[D = frac{|x - (x^2 - ln x) - 2|}{sqrt{1^2 + (-1)^2}} = frac{| -x^2 + x + ln x - 2|}{sqrt{2}}]We need to minimize the numerator ( | -x^2 + x + ln x - 2 | ). Define ( g(x) = -x^2 + x + ln x - 2 ). Since the denominator ( sqrt{2} ) is a positive constant, we focus on minimizing ( |g(x)| ).To determine if ( g(x) ) changes sign, we analyze ( h(x) = -g(x) = x^2 - x - ln x + 2 ). We find the critical points of ( h(x) ) by taking its derivative:[h'(x) = 2x - 1 - frac{1}{x}]Setting ( h'(x) = 0 ) gives the quadratic equation ( 2x^2 - x - 1 = 0 ), which has solutions ( x = 1 ) (since ( x > 0 )). Evaluating ( h(x) ) at ( x = 1 ):[h(1) = 1^2 - 1 - ln 1 + 2 = 1 - 1 - 0 + 2 = 2]The second derivative ( h''(x) = 2 + frac{1}{x^2} ) is always positive, confirming ( x = 1 ) is a minimum. Thus, ( h(x) geq 2 ) for all ( x > 0 ), meaning ( |g(x)| = h(x) ) has a minimum value of 2 at ( x = 1 ).The minimum distance is:[frac{2}{sqrt{2}} = sqrt{2}]Verifying with the point ( (1, 1) ) on the curve, the distance to the line ( y = x - 2 ) is:[frac{|1 - 1 - 2|}{sqrt{2}} = frac{2}{sqrt{2}} = sqrt{2}]Thus, the minimum distance from point ( P ) to the line is (boxed{sqrt{2}}).]

🔑:Let point P have coordinates (x, y) on the given curve. Then the derivative of y with respect to x is y' = 2x - frac{1}{x} (assuming x > 0 since the natural logarithm is defined for positive real numbers only).For the distance from the curve to the line y = x - 2 to be minimal, the curve y = x^2 - ln x must be tangent to a line parallel to y = x - 2, which has a slope of 1. Therefore, we set the derivative of the curve equal to the slope of the line:2x - frac{1}{x} = 1Solving this equation, we multiply each term by x to clear the fraction:2x^2 - 1 = xRearrange the terms to form a quadratic equation in x:2x^2 - x - 1 = 0Factoring this equation, we obtain:(x - 1)(2x + 1) = 0Since x > 0, the only acceptable solution for x is x = 1. Substituting x = 1 back into the original curve equation gives y = 1^2 - ln 1 = 1.Now we have the coordinates of the point of tangency as (1, 1). To find the distance d from this point to the line y = x - 2, we use the point-to-line distance formula:d = frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}}For the line y = x - 2, rewritten in general form, ax + by + c = 0 becomes x - y - 2 = 0. Thus, a = 1, b = -1, and c = -2:d = frac{|(1)(1) + (-1)(1) - 2|}{sqrt{1^2 + (-1)^2}} = frac{|1 - 1 - 2|}{sqrt{2}} = frac{2}{sqrt{2}} = boxed{sqrt{2}}

🧠:[<think>Okay, so I need to find the maximum value of the expression x(2 - 3x) when x is between 0 and 2/3. Hmm, let's think. This seems like a quadratic expression, right? Maybe I can expand it or put it into a standard form to find its maximum. Let me start by expanding the expression.x multiplied by (2 - 3x) would be 2x - 3x². Yeah, that's a quadratic equation in terms of x. Since the coefficient of x² is negative (-3), the parabola opens downward, which means the vertex will be the maximum point. That makes sense because if the parabola opens downward, the vertex is the highest point.So, the general form of a quadratic equation is ax² + bx + c. In this case, a is -3 and b is 2. The vertex of a parabola is at x = -b/(2a). Let me plug in the values here. So, x = -2/(2*(-3)) = -2 / (-6) = 1/3. Wait, so the x-coordinate of the vertex is 1/3. But I need to check if this x-value is within the given interval. The problem states 0 < x < 2/3. 1/3 is approximately 0.333, and 2/3 is approximately 0.666, so 1/3 is indeed between 0 and 2/3. That's good. So, the maximum occurs at x = 1/3. Now, let's find the value of the expression at x = 1/3. Plugging back into the original expression: (1/3)(2 - 3*(1/3)) = (1/3)(2 - 1) = (1/3)(1) = 1/3. So, the maximum value is 1/3.Wait, but let me double-check to make sure I didn't make any mistakes. Let me compute it again. If x = 1/3, then 2 - 3x is 2 - 3*(1/3) = 2 - 1 = 1. Then 1/3 * 1 is indeed 1/3. Okay, that seems right.Alternatively, maybe I can use calculus here? If I take the derivative of the function f(x) = 2x - 3x², the derivative f'(x) would be 2 - 6x. Setting the derivative equal to zero gives 2 - 6x = 0, so 6x = 2, which means x = 2/6 = 1/3. Then, plugging back into the original function gives the same result. So, calculus confirms that the maximum is at x = 1/3, and the maximum value is 1/3.But since the problem mentions it's an algebraic expression and doesn't specify calculus, maybe the first method using vertex formula is more appropriate. Either way, both methods lead to the same conclusion.Just to be thorough, let me check the endpoints of the interval to ensure that the maximum isn't there. Although since the vertex is inside the interval and it's a maximum, the endpoints should be lower.At x approaching 0 from the right, the expression x(2 - 3x) approaches 0*(2 - 0) = 0. At x approaching 2/3 from the left, the expression becomes (2/3)(2 - 3*(2/3)) = (2/3)(2 - 2) = (2/3)(0) = 0. So, both endpoints approach 0, which is less than 1/3. Therefore, the maximum is indeed at x = 1/3 with value 1/3.Wait, but just to make sure, is 1/3 the correct maximum? Let me also consider if there's another method. For example, completing the square. Let's try that.Starting with the expression 2x - 3x². Let me rewrite it as -3x² + 2x. Factor out the coefficient of x², which is -3:-3(x² - (2/3)x). Now, to complete the square inside the parentheses, take half of the coefficient of x, which is - (2/3) / 2 = -1/3. Square that: (-1/3)² = 1/9. So, add and subtract 1/9 inside the parentheses:-3[(x² - (2/3)x + 1/9 - 1/9)] = -3[(x - 1/3)² - 1/9] = -3(x - 1/3)² + 1/3.So, the expression can be written as -3(x - 1/3)² + 1/3. Since the square term is always non-negative, the maximum value occurs when the square term is zero, which is when x = 1/3. Then, the maximum value is 1/3. Yep, that's the same result. So, all methods confirm that the maximum is 1/3 at x = 1/3.Therefore, I think the answer is 1/3. Let me just make sure there isn't a trick or something I missed. The original interval is open, 0 < x < 2/3, but since 1/3 is inside the interval and the function is defined there, it's fine. The maximum is achieved at x = 1/3 even if the endpoints aren't included. So, everything checks out.Final AnswerThe maximum value is boxed{dfrac{1}{3}}.</think>Given (0 < x < frac{2}{3}), we need to find the maximum value of the algebraic expression (x(2 - 3x)).First, we expand the expression:[x(2 - 3x) = 2x - 3x^2]This is a quadratic expression in the form (ax^2 + bx + c) where (a = -3), (b = 2), and (c = 0). Since the coefficient of (x^2) is negative, the parabola opens downward, and the maximum value occurs at the vertex.The x-coordinate of the vertex of a quadratic (ax^2 + bx + c) is given by:[x = -frac{b}{2a}]Substituting (a = -3) and (b = 2):[x = -frac{2}{2(-3)} = frac{1}{3}]We check that (x = frac{1}{3}) is within the interval (0 < x < frac{2}{3}), which it is.Next, we substitute (x = frac{1}{3}) back into the original expression:[left(frac{1}{3}right)left(2 - 3 left(frac{1}{3}right)right) = left(frac{1}{3}right)(2 - 1) = left(frac{1}{3}right)(1) = frac{1}{3}]We also verify the endpoints of the interval. As (x) approaches 0 or (frac{2}{3}), the expression (x(2 - 3x)) approaches 0, confirming that the maximum value is indeed at (x = frac{1}{3}).Thus, the maximum value of the expression is (boxed{dfrac{1}{3}}).,<think>Okay, so I have this problem here: Given that x is between 0 and 2/3, find the maximum value of the expression x(2 - 3x). Hmm, let me think about how to approach this. First, I remember that when you have an algebraic expression and you need to find its maximum or minimum value, especially within a certain interval, calculus might be useful. Like, taking the derivative and finding critical points? But wait, maybe this can also be solved by completing the square or using vertex formula since it's a quadratic expression. Let me check.The expression is x(2 - 3x). Let me expand that first. Multiplying x by both terms inside the parentheses gives 2x - 3x². So, the expression simplifies to -3x² + 2x. Yep, that's a quadratic equation in standard form, ax² + bx + c, where a = -3, b = 2, and c = 0. Since it's a quadratic, its graph is a parabola. The coefficient of x² is negative (-3), which means the parabola opens downward. Therefore, the vertex of this parabola will be its maximum point. That makes sense because if it opens downward, the highest point is the vertex. So, the maximum value of the expression is at the vertex.Now, to find the vertex of a quadratic equation ax² + bx + c, the x-coordinate is given by -b/(2a). Let me compute that. Here, a = -3 and b = 2. So, plugging in, x = -2/(2*(-3)) = -2 / (-6) = 1/3. So, the x-coordinate of the vertex is 1/3. Now, I need to check if this x-value is within the given interval 0 < x < 2/3. Since 1/3 is approximately 0.333, and 2/3 is approximately 0.666, 1/3 is indeed between 0 and 2/3. So, the vertex is within the interval, which means the maximum value occurs at x = 1/3. Now, let's find the corresponding y-value (the maximum value of the expression). Substitute x = 1/3 into the expression x(2 - 3x). Let's compute that:First, compute 2 - 3x when x = 1/3. That's 2 - 3*(1/3) = 2 - 1 = 1. Then multiply by x, which is 1/3. So, 1*(1/3) = 1/3. Therefore, the maximum value is 1/3.Wait, let me verify that again. Maybe I should compute it using the expanded quadratic form as well to check. Original expression: -3x² + 2x. Substitute x = 1/3:-3*(1/3)^2 + 2*(1/3) = -3*(1/9) + 2/3 = -1/3 + 2/3 = 1/3. Yep, same result. So that checks out.Alternatively, using calculus: If I take the derivative of the function f(x) = x(2 - 3x) = 2x - 3x², then f'(x) = 2 - 6x. Setting the derivative equal to zero to find critical points: 2 - 6x = 0 => 6x = 2 => x = 2/6 = 1/3. So, critical point at x = 1/3. Then, to confirm it's a maximum, take the second derivative: f''(x) = -6, which is negative, confirming that the function has a maximum at x = 1/3. Therefore, substituting x = 1/3 into the original expression gives the maximum value of 1/3. But wait, just to be thorough, since the interval is open (0 < x < 2/3), we should check that x = 1/3 is indeed within the open interval and not an endpoint. Since 1/3 is approximately 0.333, which is greater than 0 and less than 2/3 (~0.666), it's inside the interval. Therefore, we don't need to check the endpoints here because the maximum is achieved inside the interval. However, if the interval had been closed, we would check the endpoints as well. But since it's open, the endpoints aren't included, but even so, the maximum is still at x =1/3 because as x approaches 0 from the right, the expression approaches 0, and as x approaches 2/3 from the left, let's see: x(2 - 3x). Let's compute the limit as x approaches 2/3 from the left. Substitute x = 2/3: 2/3*(2 - 3*(2/3)) = 2/3*(2 - 2) = 0. So as x approaches 2/3, the expression approaches 0. Similarly, approaching 0, it's also 0. Hence, the maximum indeed occurs at x =1/3 with value 1/3.Alternatively, someone might try to use completing the square to verify. Let's try that method. Starting with -3x² + 2x. Factor out the coefficient of x²: -3(x² - (2/3)x). Now, to complete the square inside the parentheses: take half of the coefficient of x, which is - (2/3)/2 = -1/3, square it: (1/3)^2 = 1/9. So add and subtract 1/9 inside the parentheses:-3[ x² - (2/3)x + 1/9 - 1/9 ] = -3[ (x - 1/3)^2 - 1/9 ] = -3(x - 1/3)^2 + 1/3.So, the expression becomes -3(x - 1/3)^2 + 1/3. Since the square term is always non-negative, the maximum value occurs when the square term is zero, i.e., when x =1/3, resulting in the maximum value of 1/3. This matches our previous results. Therefore, regardless of the method—vertex formula, calculus, or completing the square—the maximum value of x(2 - 3x) for 0 < x < 2/3 is 1/3. Just to recap the steps to make sure I didn't miss anything:1. Recognize the expression as quadratic.2. Determine the direction of the parabola (downward opening, so maximum at vertex).3. Calculate vertex x-coordinate using -b/(2a).4. Verify that the vertex is within the given interval.5. Substitute back into the original expression to find the maximum value.6. Confirm with alternative methods (calculus, completing the square) for cross-verification.All steps lead to the same conclusion. Therefore, the maximum value is 1/3. I think that covers everything. Let me just check if there's another perspective. Maybe using AM-GM inequality or something? But since this is a quadratic, those methods might not be straightforward. Let's see:The expression is x(2 - 3x). Let me denote A = x and B = 2 - 3x. Then, the product is AB. However, A and B are not necessarily positive here? Wait, x is between 0 and 2/3. So, if x > 0, then 2 - 3x would be positive when x < 2/3. Since x is between 0 and 2/3, both A and B are positive. So, maybe AM-GM could work here. AM-GM states that for non-negative numbers A and B, the product AB is maximized when A = B, but here we have AB = x(2 - 3x). Let's see if setting x = 2 - 3x gives the maximum. Solving x = 2 - 3x:x + 3x = 2 => 4x = 2 => x = 2/4 = 1/2. Wait, but we found earlier that the maximum is at x =1/3, not 1/2. Hmm, that's conflicting. So, AM-GM might not apply here because there's a coefficient involved. Let me see.Wait, AM-GM is usually for variables that are independent, but here x and (2 - 3x) are linearly dependent. So, perhaps that approach isn't directly applicable. Alternatively, if we try to set them equal, but since they have coefficients, it might not hold. For example, if we let x = k(2 - 3x) for some constant k, but that complicates things. Alternatively, maybe using substitution. Alternatively, think of the product AB where A = x and B = 2 - 3x. Let's set variables such that A and B are variables with a linear constraint. Let me define A = x and B = 2 - 3x. Then, we have a relation between A and B: B = 2 - 3A. So, the product is AB = A(2 - 3A) which is the original expression. So, to maximize AB given B = 2 - 3A. But this is again the same quadratic expression, so AM-GM isn't helpful here because there's a linear constraint between A and B. So, I think that approach doesn't lead us anywhere new. So, sticking with the quadratic methods is the right way here. Another way someone might approach this is by testing values. For example, pick x values in the interval and compute the expression. Let's try x = 1/3 first. As we saw, the expression gives 1/3. Now, try x = 0.2 (1/5). Then, x(2 - 3x) = 0.2*(2 - 0.6) = 0.2*1.4 = 0.28, which is approximately 0.28, less than 1/3 (~0.333). Try x = 0.25: 0.25*(2 - 0.75) = 0.25*1.25 = 0.3125. Still less. Try x = 0.3: 0.3*(2 - 0.9) = 0.3*1.1 = 0.33. Closer. x = 1/3 is approximately 0.333, so 0.333*(2 - 1) = 0.333*1 = 0.333. If we go beyond x =1/3, say x = 0.4: 0.4*(2 - 1.2) = 0.4*0.8 = 0.32. Less again. So, the value increases up to x =1/3 and then decreases. Therefore, confirming the maximum at x =1/3.This trial method also supports our previous conclusion. So, all methods converge to the same answer. Therefore, I feel confident that the maximum value is indeed 1/3.Final AnswerThe maximum value is boxed{dfrac{1}{3}}.</think>Given (0 < x < frac{2}{3}), we need to find the maximum value of the algebraic expression (x(2 - 3x)).First, we expand the expression:[x(2 - 3x) = 2x - 3x^2]This is a quadratic expression in the form (ax^2 + bx + c) where (a = -3), (b = 2), and (c = 0). Since the coefficient of (x^2) is negative, the parabola opens downward, indicating a maximum value at the vertex.The x-coordinate of the vertex of a quadratic (ax^2 + bx + c) is given by (-frac{b}{2a}):[x = -frac{2}{2(-3)} = frac{1}{3}]We check that (frac{1}{3}) is within the interval (0 < x < frac{2}{3}), which it is.Substituting (x = frac{1}{3}) back into the expression:[left(frac{1}{3}right)left(2 - 3left(frac{1}{3}right)right) = left(frac{1}{3}right)left(2 - 1right) = left(frac{1}{3}right)(1) = frac{1}{3}]We verify using calculus by taking the derivative of (f(x) = 2x - 3x^2):[f'(x) = 2 - 6x]Setting the derivative to zero:[2 - 6x = 0 implies x = frac{1}{3}]The second derivative (f''(x) = -6) confirms a maximum at (x = frac{1}{3}).Using completing the square:[-3x^2 + 2x = -3left(x^2 - frac{2}{3}xright) = -3left(left(x - frac{1}{3}right)^2 - frac{1}{9}right) = -3left(x - frac{1}{3}right)^2 + frac{1}{3}]The maximum value occurs when the square term is zero, resulting in (frac{1}{3}).Thus, the maximum value of the expression (x(2 - 3x)) within the interval (0 < x < frac{2}{3}) is (boxed{dfrac{1}{3}}).]

🔑:(2) When x=frac{1}{3}, the algebraic expression x(2-3x) achieves the maximum value frac{1}{3}

🧠:[<think>Okay, let's tackle this problem step by step. There are three parts, so I need to handle each one carefully. Let me start with the first part: finding ( C_n ).The problem states that we have ( n ) parallel lines given by the equations ( l_i: x - y + C_i = 0 ), where ( C_1 = sqrt{2} ) and the distance between each pair of consecutive lines increases successively by 2, 3, 4, ..., up to ( n ). Wait, actually, the problem says the distance increases "successively by 2, 3, 4, ..., ( n )." Hmm, that wording is a bit confusing. Let me parse it again.It says: "the distance between each pair of consecutive lines from these ( n ) lines increases successively by 2, 3, 4, ..., ( n )." So, starting from the first pair (between ( l_1 ) and ( l_2 )), the distance increases by 2, then between ( l_2 ) and ( l_3 ), it increases by 3, and so on, up to the last pair ( l_{n-1} ) and ( l_n ), which increases by ( n ). Wait, but the sequence 2, 3, 4, ..., ( n ) would have ( n - 1 ) terms, right? Because starting at 2 and going up to ( n ), that's ( n - 1 + 1 = n ) numbers, but the number of gaps between ( n ) lines is ( n - 1 ). Wait, actually, if there are ( n ) lines, there are ( n - 1 ) consecutive pairs. So the increments should be 2, 3, ..., up to ( n ). But that would mean starting from 2 and going up to ( n ), which is ( n - 1 ) terms? Wait, 2 to ( n ) inclusive is ( n - 1 + 1 = n ) terms. Wait, no. Wait, if you start at 2 and go to ( n ), the number of terms is ( n - 1 ), because you subtract 2 from ( n ) and add 1? Wait, let's see: from 1 to m, there are m terms. So from 2 to n, that's (n - 2 + 1) = n -1 terms. So if there are n -1 distances, each increasing by 2, 3, ..., n, but that would mean starting at 2 and ending at n, which is n -1 terms. So that matches. So the first distance (between l1 and l2) increases by 2, the next (between l2 and l3) increases by 3, and so on until the (n-1)th distance (between l_{n-1} and l_n) increases by n. Wait, but the problem says "the distance between each pair of consecutive lines ... increases successively by 2, 3, 4, ..., n". So the incremental increase in the distances is 2, 3, 4, ..., n. Wait, but if there are n-1 pairs, how can the increments go up to n? That seems conflicting. Wait, perhaps the problem is phrased as the distance between each consecutive pair starts with some initial distance, and then each next distance increases by 2, 3, etc. But maybe it's the difference between consecutive distances? Wait, the wording is a bit unclear.Wait, maybe "the distance between each pair of consecutive lines from these n lines increases successively by 2, 3, 4, ..., n." So the first distance (between l1 and l2) is 2 units, the next (l2 and l3) is 3 units, and so on, up to the (n-1)th distance (l_{n-1} and l_n) being n units. But that would require the distances themselves to be 2, 3, 4, ..., n. But the problem says "increases successively by 2, 3, 4, ..., n." Wait, the term "increases successively by" might mean that each subsequent distance is the previous one plus an increment. So if the first distance is d1, then the next distance is d1 + 2, then d2 + 3, etc. But that interpretation might not align with the problem's wording. Alternatively, maybe the differences between the constants C_i are arranged so that the distance between consecutive lines increases by 2, 3, etc. each time. Let me think.First, recall that the distance between two parallel lines ( ax + by + c_1 = 0 ) and ( ax + by + c_2 = 0 ) is given by ( frac{|c_1 - c_2|}{sqrt{a^2 + b^2}} ). In our case, the lines are ( x - y + C_i = 0 ), so ( a = 1 ), ( b = -1 ), so the distance between ( l_i ) and ( l_j ) is ( frac{|C_i - C_j|}{sqrt{1 + 1}} = frac{|C_i - C_j|}{sqrt{2}} ).Now, the distance between consecutive lines is supposed to increase successively by 2, 3, 4, ..., n. Let's parse this carefully. The problem states: "the distance between each pair of consecutive lines from these n lines increases successively by 2, 3, 4, ..., n." So starting from the first pair (l1 and l2), the distance is some value, then the next pair (l2 and l3) has a distance increased by 2, then the next (l3 and l4) increased by 3, etc. Wait, but that doesn't make much sense because "increases by" implies an additive increment. Alternatively, maybe the distances themselves form a sequence where each term is the previous term plus an increment of 2, then 3, then 4, etc. But if we have n -1 distances (since n lines have n -1 gaps), then the increments would be from 2 up to... Let's see.Wait, if the first distance is D1, then the second distance is D1 + 2, the third distance is D2 + 3 = D1 + 2 + 3, and so on. Then the k-th distance would be D1 + 2 + 3 + ... + (k+1). Wait, but if the increments start at 2, then for k = 1, increment is 2, k=2 increment is 3, up to k = n -1, increment is n. Wait, n -1 increments would start at 2 and go to n, which is n -1 terms: 2, 3, ..., n. So the total number of increments is (n -1). But 2 to n is (n -1) numbers. For example, if n = 3, then increments are 2, 3 (two increments). Wait, but n=3 lines would have two distances. If the first distance is D1, then the second distance is D1 + 2, then the third distance would be D1 + 2 + 3? Wait, no. If n=3, there are two distances. So for the first distance (between l1 and l2), maybe the increment is 2, and the next increment is 3? But that's confusing. Alternatively, perhaps the problem is that the distances themselves form an arithmetic progression with common differences increasing by 2, 3, etc. But I think I need to parse the problem again.Original problem statement: "the distance between each pair of consecutive lines from these n lines increases successively by 2, 3, 4, ..., n." So maybe the distances themselves are 2, 3, 4, ..., n? But then, for n lines, there are n -1 distances. So if n=2, one distance which is 2; if n=3, two distances: 2 and 3; if n=4, three distances: 2, 3, 4; and so on. So in general, for n lines, the distances between consecutive pairs are 2, 3, 4, ..., n. Wait, but that would require the number of distances to be n -1, so if the distances are 2, 3, ..., n, that is n -1 terms. Wait, 2 to n inclusive is (n -1) terms. For example, if n=5, distances are 2,3,4,5 (four terms). Wait, no. If n=5, the lines are l1, l2, l3, l4, l5. So four distances. But 2,3,4,5 is four numbers. So yes, if the distances between consecutive lines are 2,3,4,...,n, but n has to be such that the number of distances (n-1) equals the number of terms in the sequence from 2 to n. Wait, but the number of terms from 2 to n is n -1. So that works. Therefore, for n lines, the distances between consecutive lines are 2,3,...,n. But wait, when n=2, there is only one distance, which is 2. For n=3, two distances: 2 and 3. So in general, the k-th distance (between l_k and l_{k+1}) is 2 + (k -1). Wait, no, for k from 1 to n-1, the distance would be 2 + (k -1) = k +1. So the distance between l1 and l2 is 2, l2 and l3 is 3, ..., l_{n-1} and l_n is n. So yes, that seems to make sense.Therefore, the distance between l_i and l_{i+1} is (i +1) units. But since the lines are in the form x - y + C_i = 0, the distance between l_i and l_{i+1} is |C_{i+1} - C_i| / sqrt(2) = i +1. Therefore, |C_{i+1} - C_i| = (i +1)*sqrt(2). But since C_1 < C_2 < ... < C_n, we can drop the absolute value and write C_{i+1} - C_i = (i +1)*sqrt(2).Therefore, the difference between consecutive C_i is (i +1)*sqrt(2). Starting from C_1 = sqrt(2), we can compute C_n as C_1 + sum_{k=2}^{n} (k*sqrt(2)). Wait, let's check:Wait, the difference between C_2 and C_1 is (1 +1)*sqrt(2) = 2*sqrt(2). Therefore, C_2 = C_1 + 2*sqrt(2). Then the difference between C_3 and C_2 is (2 +1)*sqrt(2) = 3*sqrt(2), so C_3 = C_2 + 3*sqrt(2). Continuing this way, C_{i} = C_{i-1} + (i)*sqrt(2). Wait, but let's index properly. For the first difference, between i=1 and i=2, the difference is (1 +1)=2, so the multiplier is 2. Then between i=2 and i=3, it's 3. So in general, between C_{k} and C_{k+1}, the difference is (k +1)*sqrt(2). Therefore, starting from C_1, the differences are 2*sqrt(2), 3*sqrt(2), ..., up to n*sqrt(2). Wait, but for n lines, how many differences are there? For n lines, there are n -1 differences. So the differences would be for k=1 to k=n-1: (k +1)*sqrt(2). Therefore, C_n = C_1 + sum_{k=1}^{n-1} (k +1)*sqrt(2) = sqrt(2) + sqrt(2) * sum_{k=1}^{n-1} (k +1).Simplifying the sum sum_{k=1}^{n-1} (k +1) = sum_{m=2}^{n} m = [sum_{m=1}^{n} m] - 1 = [n(n +1)/2] - 1. Therefore, C_n = sqrt(2) + sqrt(2)*[n(n +1)/2 - 1] = sqrt(2)*[1 + n(n +1)/2 -1] = sqrt(2)*[n(n +1)/2]. So C_n = (sqrt(2)/2)*n(n +1). Which can be written as (n(n +1)/2)*sqrt(2). So C_n = [n(n +1)/√2]? Wait, no. sqrt(2)/2 is 1/√2. So [sqrt(2)/2] *n(n +1) = [n(n +1)/2] * sqrt(2). Either form is acceptable, but perhaps rationalizing, it's (n(n +1)√2)/2. So maybe better written as (frac{n(n+1)}{2} sqrt{2}). Therefore, that's the value of C_n.Wait, let's verify this with a small n. For example, n=2. Then C_2 should be C_1 + 2*sqrt(2) = sqrt(2) + 2*sqrt(2) = 3*sqrt(2). According to the formula: (2*3/2)*sqrt(2) = 3*sqrt(2). Correct. For n=3: C_3 = C_2 + 3*sqrt(2) = 3*sqrt(2) + 3*sqrt(2) = 6*sqrt(2). According to formula: (3*4/2)*sqrt(2) = 6*sqrt(2). Correct. Similarly, n=4: (4*5/2)*sqrt(2) = 10*sqrt(2). Which would be built as C_1 +2+3+4 each multiplied by sqrt(2): sqrt(2) +2√2 +3√2 +4√2 = (1+2+3+4)√2 =10√2. Correct. So the formula seems to hold. Therefore, the first part answer is C_n = [n(n +1)/2]√2.Moving to part 2: Find the area of the shape formed by l_n : x - y + C_n = 0 and the x-axis and y-axis. So we need to find the area of the triangle formed by the line l_n, the x-axis, and the y-axis. To find this area, we can find the x-intercept and y-intercept of the line l_n, then compute the area as 1/2 * |x-intercept| * |y-intercept|.The line l_n is x - y + C_n = 0. Let's rearrange it to intercept form. Solving for x when y=0: x + C_n = 0 => x-intercept = -C_n. Solving for y when x=0: -y + C_n = 0 => y-intercept = C_n. Therefore, the x-intercept is (-C_n, 0) and the y-intercept is (0, C_n). The area is 1/2 * | -C_n | * | C_n | = 1/2 * C_n^2.Given that C_n = [n(n +1)/2]√2, so C_n^2 = [n^2(n +1)^2 /4] * 2 = [n^2(n +1)^2 /2]. Therefore, the area is 1/2 * [n^2(n +1)^2 /2] = [n^2(n +1)^2]/4. Wait, let me verify:Wait, C_n = (n(n +1)/2)√2. Then C_n squared is [n(n+1)/2]^2 * 2. So [n²(n+1)² /4] * 2 = [n²(n+1)² /2]. Then the area is (1/2)*[n²(n+1)² /2] = n²(n+1)² /4. Wait, but hold on: x-intercept is -C_n, y-intercept is C_n. So the area is (1/2)*|x-intercept|*|y-intercept| = (1/2)*C_n * C_n = (1/2)*C_n². Therefore, substituting C_n² = [n(n+1)/2]^2 * 2, which is [n²(n+1)² /4]*2 = n²(n+1)² /2. Therefore, the area is (1/2)*(n²(n+1)² /2) = n²(n+1)² /4. So that's correct.Alternatively, let's take n=2. Then C_n = 3√2. The line x - y + 3√2 =0. x-intercept is -3√2, y-intercept is 3√2. Area is 1/2 * 3√2 *3√2 = 1/2 *9*2=9. According to the formula: 2²*3² /4 = 4*9 /4=9. Correct. For n=3, C_n=6√2. Area would be 1/2*(6√2)^2=1/2*72=36. The formula: 3²*4² /4=9*16 /4=144/4=36. Correct. So the formula holds. Therefore, the answer is ( frac{n^2(n+1)^2}{4} ).Now part 3: Find the area of the shape formed by ( l_{n-1} : x - y + C_{n-1} = 0 ), ( l_n : x - y + C_n = 0 ), and the x-axis and y-axis. So this is the area between the lines l_{n-1}, l_n, the x-axis, and the y-axis. To compute this, we need to find the region bounded by these four lines. However, since l_{n-1} and l_n are parallel, the shape formed might be a quadrilateral, but since they intersect the axes, it's actually a trapezoid or a triangle. Wait, but two parallel lines and the two axes. Let me visualize.If we have two parallel lines, both intersecting the x-axis and y-axis, then the area between them and the axes would form a quadrilateral. However, depending on the positions of the lines, the figure could be a trapezoid. Alternatively, if one of the lines is closer to the origin, the area might be a quadrilateral bounded by the two lines and the two axes. Let me find the intercepts for both lines.For l_{n}: x - y + C_n =0. As before, x-intercept is -C_n, y-intercept is C_n.For l_{n-1}: x - y + C_{n-1} =0. Similarly, x-intercept is -C_{n-1}, y-intercept is C_{n-1}.Since C_{n-1} < C_n (given that C_1 < C_2 < ... < C_n), the line l_n is farther from the origin compared to l_{n-1}. So when we plot both lines, the x-intercept of l_n is more negative (farther left on the x-axis), and the y-intercept is higher up on the y-axis. The region bounded by these two lines and the axes would be the area between the two lines and the axes, which is a quadrilateral with vertices at (-C_n, 0), (0, C_n), (0, C_{n-1}), and (-C_{n-1}, 0). Wait, is that correct? Let me sketch mentally.The line l_n intersects the x-axis at (-C_n, 0) and the y-axis at (0, C_n). The line l_{n-1} intersects the x-axis at (-C_{n-1}, 0) and the y-axis at (0, C_{n-1}). So the quadrilateral is formed by connecting these four points: from (-C_n, 0) along l_n to (0, C_n), then down along the y-axis to (0, C_{n-1}), then along l_{n-1} to (-C_{n-1}, 0), and back along the x-axis to (-C_n, 0). So it's a trapezoid with two sides on the axes and the other two sides being parts of the lines l_n and l_{n-1}.Alternatively, since the lines are parallel, the distance between them is constant, but in this case, the area can be found by subtracting the area under l_{n-1} from the area under l_n. But since the area under each line with the axes is a triangle, the area between them would be the difference between the two triangle areas.But wait, the quadrilateral described is actually the difference between the two triangles. Let me check:The area formed by l_n and the axes is a triangle with area 1/2 * C_n^2 as found in part 2. Similarly, the area formed by l_{n-1} and the axes is 1/2 * C_{n-1}^2. Therefore, the area between them would be the difference: 1/2*(C_n^2 - C_{n-1}^2). However, this might not be correct because the region bounded by both lines and both axes is not simply the difference of the two triangles. Wait, actually, if you consider the quadrilateral, it's a trapezoid with two parallel sides (the segments on l_n and l_{n-1} between the axes) and the legs on the axes. The area of a trapezoid is the average of the two parallel sides multiplied by the distance between them. However, in this case, the two parallel sides are along the lines l_n and l_{n-1}, and the distance between them is the distance between the lines l_n and l_{n-1}.But perhaps there's a simpler way. Let's compute the coordinates of the four vertices of the quadrilateral:1. Intersection of l_n with x-axis: (-C_n, 0)2. Intersection of l_n with y-axis: (0, C_n)3. Intersection of l_{n-1} with y-axis: (0, C_{n-1})4. Intersection of l_{n-1} with x-axis: (-C_{n-1}, 0)So the quadrilateral is a trapezoid with vertices at (-C_n, 0), (0, C_n), (0, C_{n-1}), (-C_{n-1}, 0). To find the area, we can use the shoelace formula.Alternatively, note that this is a trapezoid with the two parallel sides being the segments from (-C_n, 0) to (0, C_n) (which is the line l_n) and from (-C_{n-1}, 0) to (0, C_{n-1}) (which is the line l_{n-1}). However, in trapezoid terms, the two parallel sides are the ones between the two lines. Wait, actually, no. In a trapezoid, the two parallel sides are the ones that are the "bases". Here, the two bases would be the parts of the lines l_n and l_{n-1} between the axes. But since the lines are not horizontal or vertical, their segments between the axes are not parallel. Wait, actually, the lines l_n and l_{n-1} are parallel to each other (they have the same slope), so the segments connecting their intercepts are not parallel. Wait, this is confusing. Wait, the lines l_n and l_{n-1} are both of the form x - y + C =0, so they are parallel with slope 1. Therefore, the segments connecting their intercepts are the lines themselves, but between the axes. However, the two lines l_n and l_{n-1} are parallel, but the sides of the quadrilateral are the two lines between their intercepts, and the two axes. Therefore, the two non-parallel sides of the trapezoid are along the x-axis and y-axis. Wait, but the x-axis and y-axis are perpendicular, so the figure is actually a trapezoid with one pair of sides parallel (the two lines l_n and l_{n-1}) and the other two sides along the axes.However, the distance between the lines l_n and l_{n-1} is |C_n - C_{n-1}| / sqrt(2). But the height of the trapezoid (the distance between the two parallel sides) would be this distance. However, the area of a trapezoid is (a + b)/2 * h, where a and b are the lengths of the two parallel sides, and h is the distance between them.First, let's compute the lengths of the two parallel sides. The length of the segment of l_n between the axes: from (-C_n, 0) to (0, C_n). The length is sqrt( (C_n - 0)^2 + (0 - (-C_n))^2 ) = sqrt( C_n^2 + C_n^2 ) = sqrt(2 C_n^2 ) = C_n sqrt(2). Similarly, the length of the segment of l_{n-1} between the axes is C_{n-1} sqrt(2). So the two parallel sides have lengths a = C_{n-1} sqrt(2) and b = C_n sqrt(2). The distance h between the two lines is |C_n - C_{n-1}| / sqrt(2). But from part 1, we know that C_n = C_{n-1} + n sqrt(2). Therefore, |C_n - C_{n-1}| = n sqrt(2). Therefore, h = n sqrt(2) / sqrt(2) = n. Therefore, the area would be (a + b)/2 * h = (C_{n-1} sqrt(2) + C_n sqrt(2))/2 * n = sqrt(2)/2 (C_{n-1} + C_n) * n.Alternatively, since C_n = C_{n-1} + n sqrt(2), then C_{n} + C_{n-1} = 2 C_{n-1} + n sqrt(2). But maybe there's a simpler expression. Let's compute C_{n} and C_{n-1} using the formula from part 1. From part 1, C_k = [k(k +1)/2] sqrt(2). Therefore, C_{n} = [n(n +1)/2] sqrt(2), and C_{n-1} = [(n -1)n /2] sqrt(2). Therefore, C_n + C_{n-1} = [n(n +1)/2 + n(n -1)/2] sqrt(2) = [ (n^2 +n +n^2 -n)/2 ] sqrt(2) = [ (2n^2)/2 ] sqrt(2) = n^2 sqrt(2). Therefore, sqrt(2)/2 * (C_{n} + C_{n-1}) * n = sqrt(2)/2 * n^2 sqrt(2) * n = (sqrt(2) * sqrt(2))/2 * n^3 = (2/2) n^3 = n^3. Therefore, the area is n^3.But let me verify this with an example. Let's take n=2. Then C_2 = 3 sqrt(2), C_1 = sqrt(2). The area should be the area between l_1 and l_2 and the axes. The quadrilateral has vertices at (-3√2, 0), (0, 3√2), (0, √2), (-√2, 0). Using the shoelace formula:Coordinates:1. (-3√2, 0)2. (0, 3√2)3. (0, √2)4. (-√2, 0)Back to (-3√2, 0)Compute the shoelace sum:Sum1 = (-3√2 * 3√2) + (0 * √2) + (0 * 0) + (-√2 * 0) = (-9*2) + 0 + 0 + 0 = -18Sum2 = (0 * 0) + (3√2 * 0) + (√2 * -√2) + (0 * -3√2) = 0 + 0 + (-2) + 0 = -2Area = |Sum1 - Sum2| / 2 = |-18 - (-2)| /2 = |-16| /2 = 8.According to the formula n^3 = 2^3 =8. Correct. For n=3, the area should be 27. Let's check:C_3 = 6√2, C_2 = 3√2.Quadrilateral vertices:(-6√2, 0), (0, 6√2), (0, 3√2), (-3√2, 0).Shoelace formula:Sum1 = (-6√2 *6√2) + (0 *3√2) + (0 *0) + (-3√2 *0) = (-36*2) +0 +0 +0= -72Sum2 = (0*0) + (6√2*0) + (3√2*(-3√2)) + (0*(-6√2)) =0 +0 + (-9*2) +0= -18Area= |-72 - (-18)| /2 = |-54| /2=27. Which matches n^3=27. So the formula holds.Therefore, the area is n^3.Therefore, the answers are:1. ( C_n = frac{n(n+1)}{2} sqrt{2} )2. Area = ( frac{n^2(n+1)^2}{4} )3. Area = ( n^3 )Final Answer1. (boxed{dfrac{n(n + 1)}{2} sqrt{2}})2. (boxed{dfrac{n^2(n + 1)^2}{4}})3. (boxed{n^3})</think>Given ( n ) parallel lines ( l_i: x - y + C_i = 0 ) where ( C_1 = sqrt{2} ) and the distance between each pair of consecutive lines increases successively by 2, 3, 4, ..., ( n ):1. Finding ( C_n ): The distance between consecutive lines ( l_i ) and ( l_{i+1} ) is given by ( frac{|C_{i+1} - C_i|}{sqrt{2}} ). This distance increases by ( i+1 ) units for each consecutive pair. Therefore, the difference ( C_{i+1} - C_i = (i+1)sqrt{2} ). Summing these differences from ( C_1 ) to ( C_n ): [ C_n = C_1 + sum_{k=2}^{n} ksqrt{2} = sqrt{2} + sqrt{2} sum_{k=2}^{n} k ] Simplifying the sum: [ sum_{k=2}^{n} k = sum_{k=1}^{n} k - 1 = frac{n(n+1)}{2} - 1 ] Therefore, [ C_n = sqrt{2} left( 1 + frac{n(n+1)}{2} - 1 right) = sqrt{2} cdot frac{n(n+1)}{2} ] Thus, [ C_n = boxed{dfrac{n(n + 1)}{2} sqrt{2}} ]2. Area of the shape formed by ( l_n ), the x-axis, and the y-axis: The line ( l_n ) intersects the x-axis at ( (-C_n, 0) ) and the y-axis at ( (0, C_n) ). The area of the triangle formed is: [ text{Area} = frac{1}{2} times |xtext{-intercept}| times |ytext{-intercept}| = frac{1}{2} times C_n times C_n = frac{1}{2} C_n^2 ] Substituting ( C_n = frac{n(n+1)}{2} sqrt{2} ): [ C_n^2 = left( frac{n(n+1)}{2} sqrt{2} right)^2 = frac{n^2(n+1)^2}{4} times 2 = frac{n^2(n+1)^2}{2} ] Therefore, [ text{Area} = frac{1}{2} times frac{n^2(n+1)^2}{2} = frac{n^2(n+1)^2}{4} ] Thus, [ text{Area} = boxed{dfrac{n^2(n + 1)^2}{4}} ]3. Area of the shape formed by ( l_{n-1} ), ( l_n ), the x-axis, and the y-axis: The area between the lines ( l_{n-1} ) and ( l_n ) and the axes forms a trapezoid. The vertices of this trapezoid are ( (-C_n, 0) ), ( (0, C_n) ), ( (0, C_{n-1}) ), and ( (-C_{n-1}, 0) ). The area can be calculated using the formula for the area of a trapezoid: [ text{Area} = frac{1}{2} times (C_n + C_{n-1}) times n ] Substituting ( C_n = frac{n(n+1)}{2} sqrt{2} ) and ( C_{n-1} = frac{(n-1)n}{2} sqrt{2} ): [ C_n + C_{n-1} = frac{n(n+1)}{2} sqrt{2} + frac{(n-1)n}{2} sqrt{2} = frac{n^2 sqrt{2}}{2} (n+1 + n - 1) = n^2 sqrt{2} ] Therefore, [ text{Area} = frac{1}{2} times n^2 sqrt{2} times n times frac{1}{sqrt{2}} = n^3 ] Thus, [ text{Area} = boxed{n^3} ],<think>Alright, let me try to tackle this problem step by step. It's about parallel lines and distances between them, and then areas formed with the axes. Let me break it down part by part.First, the problem states that we have n parallel lines given by the equations l₁: x - y + C₁ = 0, with C₁ = √2, and then l₂, l₃, ..., lₙ with constants C₂, C₃, ..., Cₙ where each subsequent C is larger than the previous (C₁ < C₂ < C₃ < ... < Cₙ). The key part here is that the distance between each pair of consecutive lines increases successively by 2, 3, 4, ..., n. Wait, actually, hold on. Let me parse that again."The distance between each pair of consecutive lines from these n lines increases successively by 2, 3, 4, ..., n."Hmm. So starting from the first pair (l₁ and l₂), the distance between them is 2 units. Then the distance between l₂ and l₃ is 3 units more than the previous? Wait, no. If it's increasing by 2, 3, 4, ..., n, maybe the increments are added each time? Wait, maybe the distances themselves form a sequence where each term increases by 2, 3, 4, etc. Wait, perhaps I need to re-read that.Original problem says: "the distance between each pair of consecutive lines from these n lines increases successively by 2, 3, 4, ..., n." Wait, but there are n lines, so there are (n-1) gaps between them. So the increase in distance is by 2, 3, 4, ..., but how does that sequence go?Wait, maybe the distance between the first pair (l₁ and l₂) is 2, then between l₂ and l₃ is 3, then l₃ and l₄ is 4, etc., up to the (n-1)th distance being n. But 2, 3, 4, ..., n would give (n - 1) terms if starting from 2? Wait, 2 to n inclusive is (n - 1 + 1) numbers? Wait, no. From 2 to n, the number of terms is n - 1. For example, if n=5, then 2,3,4,5: that's 4 terms, which is n-1. So if there are n lines, there are n-1 distances between consecutive lines, and these distances are 2, 3, 4, ..., n. Wait, but 2,3,4,...,n has n-1 terms only if starting at 2 and ending at n. But for example, if n=3, then the distances would be 2 and 3, which are two distances (n-1=2). So that seems to fit. So the distances between consecutive lines start at 2 and each subsequent distance increases by 1. So the first distance is 2, next is 3, then 4, etc., up to the (n-1)th distance being n. So each distance is 2,3,4,...,n. Therefore, the sequence of distances is 2,3,4,...,n for the (n-1) gaps. Wait, but 2 to n inclusive is (n - 1) numbers if you start at 2? Wait, 2 to n is (n - 1) numbers only if starting at 1. Wait, actually, from 2 to n inclusive, how many terms are there? It's (n - 2 + 1) = n - 1 terms. For example, n=3: 2,3. That's 2 terms. Yes, so n-1 terms. So the distances between consecutive lines are 2,3,4,...,n. So distance between l₁ and l₂ is 2, l₂ and l₃ is 3, ..., l_{n-1} and l_n is n. So that's the key point.Given that, the first part asks to find C_n.So, since the lines are parallel, the difference between their constants C_i relates to the distance between them. The formula for the distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is |c₁ - c₂| / sqrt(a² + b²). In our case, the lines are x - y + C_i = 0, so a=1, b=-1. Therefore, the distance between l_i and l_j is |C_i - C_j| / sqrt(1 + 1) = |C_i - C_j| / sqrt(2).Given that the distance between l₁ and l₂ is 2, so |C₂ - C₁| / sqrt(2) = 2. Since C₂ > C₁, we can write (C₂ - C₁) = 2 * sqrt(2). Similarly, the distance between l₂ and l₃ is 3, so (C₃ - C₂) = 3 * sqrt(2). Continuing this way, each consecutive distance is (k+1)*sqrt(2), where k starts at 1? Wait, the first distance is 2, then 3, etc., so for the distance between l₁ and l₂, it's 2, which would correspond to k=2. Hmm, but perhaps it's better to note that each distance between l_i and l_{i+1} is (i+1)*sqrt(2), starting from i=1 to i=n-1.Wait, no. Wait, the first gap (l₁ to l₂) is distance 2, so that would correspond to (C₂ - C₁) = 2*sqrt(2). Then the next gap (l₂ to l₃) is distance 3, so (C₃ - C₂) = 3*sqrt(2). So in general, the difference between C_{k+1} and C_k is (k+1)*sqrt(2). Wait, but starting from k=1, where k=1 gives C₂ - C₁ = 2*sqrt(2), k=2 gives C₃ - C₂ = 3*sqrt(2), ..., up to k = n-1, which gives C_n - C_{n-1} = n*sqrt(2). Therefore, the total difference from C₁ to C_n would be the sum from k=2 to k=n of k*sqrt(2). Wait, but the differences are 2*sqrt(2), 3*sqrt(2), ..., n*sqrt(2). So sum_{i=2}^n i*sqrt(2) = sqrt(2)*(sum_{i=2}^n i) = sqrt(2)*(sum_{i=1}^n i - 1) = sqrt(2)*(n(n+1)/2 - 1). Wait, but hold on, the total difference C_n - C₁ is the sum of all the individual differences: (C₂ - C₁) + (C₃ - C₂) + ... + (C_n - C_{n-1}) ) = C_n - C₁. So this sum is (2 + 3 + ... + n)*sqrt(2). Therefore, C_n = C₁ + sqrt(2)*(2 + 3 + ... + n). Since C₁ is given as sqrt(2), then C_n = sqrt(2) + sqrt(2)*(2 + 3 + ... + n) = sqrt(2)*(1 + 2 + 3 + ... + n - 1 + n - 1)? Wait, no. Wait, 2 + 3 + ... + n is equal to (sum from 1 to n) minus 1. So sum_{k=2}^n k = sum_{k=1}^n k - 1 = n(n+1)/2 - 1. Therefore, C_n = sqrt(2) + sqrt(2)*(n(n+1)/2 - 1) = sqrt(2)*(1 + n(n+1)/2 - 1) = sqrt(2)*(n(n+1)/2). Wait, but wait, hold on:Wait, if C_n - C₁ = sum_{k=2}^n (k*sqrt(2)) = sqrt(2)*(sum_{k=2}^n k). Therefore, C_n = C₁ + sqrt(2)*(sum_{k=2}^n k). Since sum_{k=2}^n k = sum_{k=1}^n k - 1 = [n(n+1)/2] - 1. Therefore, C_n = sqrt(2) + sqrt(2)*[n(n+1)/2 - 1] = sqrt(2)*(1 + n(n+1)/2 - 1) = sqrt(2)*(n(n+1)/2). Wait, that seems a bit confusing. Let me check again.Wait, if the differences between consecutive Cs are (C₂ - C₁) = 2√2, (C₃ - C₂) = 3√2, ..., (C_n - C_{n-1}) = n√2. Therefore, the total difference C_n - C₁ is 2√2 + 3√2 + ... + n√2. Factor out √2: √2*(2 + 3 + ... + n). Then C_n = C₁ + √2*(sum from k=2 to k=n of k). Since C₁ = √2, then:C_n = √2 + √2*(sum from 2 to n of k) = √2*(1 + sum from 2 to n of k). But sum from 2 to n of k is equal to sum from 1 to n of k minus 1. So sum from 2 to n of k = [n(n+1)/2] - 1. Therefore:C_n = √2*(1 + [n(n+1)/2 - 1]) = √2*(n(n+1)/2). So this simplifies to C_n = (√2)*(n(n+1)/2). Therefore, the answer for part 1 is C_n = (n(n+1)/2) * √2.But wait, let me confirm this with a small n. Let's take n=2. Then there are two lines, l₁ and l₂. The distance between them should be 2. Let's compute C₂. According to the formula, C₂ = (2*3/2)*√2 = 3√2. Then the distance between l₁ and l₂ is |C₂ - C₁| / √2 = |3√2 - √2| / √2 = (2√2)/√2 = 2. Correct. For n=3, the lines are l₁, l₂, l₃. The distances between them are 2 and 3. So C₂ = C₁ + 2√2 = √2 + 2√2 = 3√2. Then C₃ = C₂ + 3√2 = 3√2 + 3√2 = 6√2. According to the formula, C₃ = (3*4/2)*√2 = 6√2. Correct. So the formula seems to hold. So part 1's answer is C_n = [n(n+1)/2]√2.Okay, moving on to part 2: Find the area of the shape formed by lₙ: x - y + Cₙ = 0 and the x-axis and y-axis.So we need to find the area bounded by the line lₙ, the x-axis (y=0), and the y-axis (x=0). The shape formed is a triangle. To find the area, we can find the intercepts of the line lₙ with the x-axis and y-axis, then compute the area of the triangle.First, let's find the x-intercept of lₙ. Set y=0 in the equation x - y + Cₙ = 0: x + Cₙ = 0 => x = -Cₙ. Similarly, the y-intercept is found by setting x=0: -y + Cₙ = 0 => y = Cₙ.Therefore, the intercepts are at (-Cₙ, 0) and (0, Cₙ). The area of the triangle formed by these two intercepts and the origin is (1/2)*base*height. The base is the distance along the x-axis from 0 to -Cₙ, which is | -Cₙ | = Cₙ. Similarly, the height is Cₙ. So the area is (1/2)*Cₙ*Cₙ = (1/2)Cₙ².But we need to express this in terms of n. From part 1, we know Cₙ = [n(n+1)/2]√2. Therefore, Cₙ² = [n²(n+1)² / 4] * 2 = [n²(n+1)² / 2]. Therefore, the area is (1/2)*[n²(n+1)² / 2] = [n²(n+1)²] / 4.Wait, let me check this again. If Cₙ = (n(n+1)/2)√2, then Cₙ² = [n(n+1)/2]^2 * 2. Then (1/2)*Cₙ² = (1/2)*[n²(n+1)²/4]*2 = (1/2)*(n²(n+1)²/2)*2 = (1/2)*(n²(n+1)²). Wait, no, perhaps I miscalculated.Wait, Cₙ = (n(n+1)/2)√2. So Cₙ squared is [n(n+1)/2]^2 * (√2)^2 = [n²(n+1)²/4] * 2 = [n²(n+1)²/2]. Then the area is (1/2)*Cₙ² = (1/2)*(n²(n+1)²/2) = n²(n+1)² / 4. So the area is (n²(n+1)²)/4. Hmm, that seems large, but let's check with n=1. Wait, but n must be at least 1, but in the problem statement, we have n lines starting with C₁, so n is at least 1. If n=1, then part 2 is the area formed by l₁ and the axes. Let's compute that.For n=1, C₁=√2. The line is x - y + √2 = 0. The x-intercept is x = -√2, y=0. The y-intercept is y=√2, x=0. The area is (1/2)*√2*√2 = (1/2)*2 = 1. According to the formula, (1²*2²)/4 = (1*4)/4 = 1. Correct. For n=2, C₂=3√2. Then the area is (1/2)*(3√2)^2 = (1/2)*9*2 = 9. According to the formula, (2²*3²)/4 = (4*9)/4 = 9. Correct. For n=3, C₃=6√2. Area is (1/2)*(6√2)^2 = (1/2)*36*2 = 36. Formula: (3²*4²)/4 = (9*16)/4 = 144/4 = 36. Correct. So the formula seems to hold. So the answer for part 2 is n²(n+1)² / 4.Wait, but hang on. Let me confirm again with n=2. The line is x - y + 3√2 = 0. The x-intercept is at x = -3√2, y-intercept at y=3√2. Area is (1/2)*(3√2)*(3√2) = (1/2)*(9*2) = 9. Which matches n²(n+1)² /4 for n=2: 4*9 /4 = 9. Correct. So part 2's answer is n²(n+1)² / 4.Moving on to part 3: Find the area of the shape formed by l_{n-1}: x - y + C_{n-1} = 0, l_n: x - y + C_n = 0, and the x-axis and y-axis.So now, instead of just one line and the axes, we have two lines (l_{n-1} and l_n) and the axes forming a shape. The area would be the region bounded by these four: l_{n-1}, l_n, x-axis, and y-axis. Let me visualize this.First, each of the lines l_{n-1} and l_n will intercept the x-axis and y-axis at some points. The shape formed by these two lines and the axes is likely a quadrilateral, but we need to check the exact intercepts.Alternatively, perhaps it's a trapezoid or some other figure. Let's find the intercepts for both lines.For l_{n}: x - y + C_n = 0, as before, x-intercept is (-C_n, 0), y-intercept is (0, C_n).For l_{n-1}: x - y + C_{n-1} = 0, similarly, x-intercept is (-C_{n-1}, 0), y-intercept is (0, C_{n-1}).Now, the figure formed by these two lines and the axes would be a quadrilateral with vertices at (-C_n, 0), (0, C_n), (0, C_{n-1}), and (-C_{n-1}, 0). Wait, is that correct? Let me think.The two lines l_n and l_{n-1} intersect the x-axis at (-C_n, 0) and (-C_{n-1}, 0), respectively. They intersect the y-axis at (0, C_n) and (0, C_{n-1}). The quadrilateral is formed by connecting these four points. However, the lines l_n and l_{n-1} themselves intersect each other only at infinity since they are parallel. Therefore, the figure is actually a trapezoid with vertices at (-C_n, 0), (0, C_n), (0, C_{n-1}), (-C_{n-1}, 0). But wait, actually, the lines are parallel, so the sides connecting (-C_n, 0) to (-C_{n-1}, 0) along the x-axis, and (0, C_n) to (0, C_{n-1}) along the y-axis. The other two sides are the segments of l_n and l_{n-1} between the axes. But since the lines are parallel, those two sides are parallel. Therefore, the shape is a trapezoid.The area of a trapezoid is given by the average of the two parallel sides multiplied by the distance between them. However, in this case, the trapezoid is not between two parallel lines, but between two pairs of axes. Wait, maybe a better approach is to compute the area by subtracting the area of the smaller triangle (formed by l_{n-1} and the axes) from the larger triangle (formed by l_n and the axes). Then the area between l_{n-1}, l_n, and the axes would be the difference between the two triangle areas.So the area would be Area(l_n) - Area(l_{n-1}).From part 2, we know that the area formed by l_k and the axes is (k²(k+1)²)/4. Therefore, substituting k = n and k = n-1:Area = (n²(n+1)²)/4 - ((n-1)²n²)/4.Factor out 1/4 and n²:= (n²/4)[(n+1)² - (n-1)²].Expand (n+1)² - (n-1)²:= [n² + 2n + 1] - [n² - 2n + 1] = 4n.Therefore, Area = (n²/4)(4n) = n³.Wait, that's interesting. So the area between l_{n-1}, l_n, and the axes is n³.Let me verify this with an example. Take n=2. Then l₂ has C₂ = 3√2, and l₁ has C₁ = √2. The area formed by l₂ and the axes is (2²*3²)/4 = 36/4 = 9. The area formed by l₁ and the axes is (1²*2²)/4 = 4/4 = 1. Then the difference is 9 - 1 = 8. According to the formula, n³ = 2³ = 8. Correct.Another example, n=3. l₃: C₃=6√2, area with axes is (3²*4²)/4 = 144/4 = 36. l₂: C₂=3√2, area is (2²*3²)/4= 36/4=9. Difference is 36 - 9 = 27. n³=3³=27. Correct. So this formula seems to hold.Therefore, part 3's answer is n³.But let me think again to be thorough. The area between l_{n} and the axes is A_n = (n²(n+1)²)/4. The area between l_{n-1} and the axes is A_{n-1} = ((n-1)²n²)/4. Then the difference is A_n - A_{n-1} = [n²(n+1)² - (n-1)²n²]/4 = n²[(n+1)² - (n-1)²]/4. As computed before, this is n²*(4n)/4 = n³. Therefore, correct.Alternatively, if I consider the trapezoid area formula: the two parallel sides are the segments of l_n and l_{n-1} between the axes. The distance between these two lines is given as n (from part 1's distance sequence). Wait, but in the problem statement, the distance between l_{n-1} and l_n is n. Wait, but according to the distance formula, the distance between l_{n-1} and l_n is |C_n - C_{n-1}| / sqrt(2) = (C_n - C_{n-1}) / sqrt(2) = n*sqrt(2)/sqrt(2) = n. So the distance between the two lines is n. However, when computing the area of the trapezoid, the formula is (sum of the lengths of the two parallel sides) multiplied by the distance between them, divided by 2. Wait, but in this case, the two "parallel sides" would be the segments along l_n and l_{n-1} between the axes. Wait, but those segments are not straight line sides; the trapezoid is formed by the intercepts. Wait, perhaps I need to re-examine the figure.Alternatively, since the two lines are parallel and the trapezoid is formed by their intercepts with the axes, the area can be found by the difference between the two triangles. As we saw earlier, the difference is indeed n³. So the formula holds. Therefore, part 3's answer is n³.So summarizing:1. C_n = [n(n+1)/2]√2.2. Area with axes: n²(n+1)² /4.3. Area between l_{n-1}, l_n, and axes: n³.But let me check for n=1. Wait, n=1, part 3 would require l₀, which doesn't exist. So n must be at least 2. The problem doesn't specify constraints on n, but since part 3 mentions l_{n-1}, n must be at least 2. So the answer for part 3 is valid for n≥2. But the problem might not have such constraints.In any case, based on the calculations, these are the answers.Final Answer1. boxed{C_n = dfrac{n(n + 1)}{2} sqrt{2}}2. boxed{dfrac{n^2 (n + 1)^2}{4}}3. boxed{n^3}</think>Given ( n ) parallel lines ( l_1: x - y + C_1 = 0 ) with ( C_1 = sqrt{2} ), ( l_2: x - y + C_2 = 0 ), ( l_3: x - y + C_3 = 0 ), ..., ( l_n: x - y + C_n = 0 ) (where ( C_1 < C_2 < C_3 < ldots < C_n )), the distance between each pair of consecutive lines increases successively by 2, 3, 4, ..., ( n ).1. Finding ( C_n ): - The distance between consecutive lines ( l_i ) and ( l_{i+1} ) is ( (i+1) sqrt{2} ). - The total difference between ( C_n ) and ( C_1 ) is the sum of these distances: [ C_n - C_1 = sqrt{2} sum_{k=2}^{n} k ] - The sum ( sum_{k=2}^{n} k ) is equal to ( sum_{k=1}^{n} k - 1 = frac{n(n+1)}{2} - 1 ). - Therefore, [ C_n = C_1 + sqrt{2} left( frac{n(n+1)}{2} - 1 + 1 right) = sqrt{2} left( frac{n(n+1)}{2} right) ] - Hence, [ C_n = boxed{dfrac{n(n + 1)}{2} sqrt{2}} ]2. Finding the area of the shape formed by ( l_n ) and the axes: - The x-intercept of ( l_n ) is ( -C_n ) and the y-intercept is ( C_n ). - The area of the triangle formed is: [ text{Area} = frac{1}{2} times | -C_n | times | C_n | = frac{1}{2} C_n^2 ] - Substituting ( C_n = frac{n(n+1)}{2} sqrt{2} ): [ text{Area} = frac{1}{2} left( frac{n(n+1)}{2} sqrt{2} right)^2 = frac{1}{2} left( frac{n^2(n+1)^2}{4} times 2 right) = frac{n^2(n+1)^2}{4} ] - Hence, [ boxed{dfrac{n^2 (n + 1)^2}{4}} ]3. Finding the area of the shape formed by ( l_{n-1} ), ( l_n ), and the axes: - The area is the difference between the areas formed by ( l_n ) and ( l_{n-1} ) with the axes: [ text{Area} = text{Area}(l_n) - text{Area}(l_{n-1}) ] - Using the result from part 2: [ text{Area} = frac{n^2(n+1)^2}{4} - frac{(n-1)^2 n^2}{4} = frac{n^2}{4} left( (n+1)^2 - (n-1)^2 right) ] - Simplifying: [ (n+1)^2 - (n-1)^2 = 4n ] - Therefore, [ text{Area} = frac{n^2}{4} times 4n = n^3 ] - Hence, [ boxed{n^3} ]]

🔑:1. The distance from the origin O to l_1 is 1, from O to l_2 is 1 + 2, and from O to l_n is d_n = 1 + 2 + ldots + n = frac{n(n+1)}{2}.Since C_n = sqrt{2} d_n,we have C_n = frac{sqrt{2}n(n+1)}{2}.2. Let the line l_n : x - y + C_n = 0 intersect the x-axis at M and the y-axis at N. The area of triangle triangle OMN isS_{triangle OMN} = frac{1}{2} |OM| cdot |ON| = frac{1}{2} C_n^2 = frac{n^2(n+1)^2}{4}.3. The shape formed is an isosceles trapezoid. From (2), we know that S_n = frac{n^2(n+1)^2}{4}. Therefore, S_{n-1} = frac{(n-1)^2 n^2}{4}.So, S_n - S_{n-1} = frac{n^2(n+1)^2}{4} - frac{(n-1)^2 n^2}{4} = n^3.Hence, the requested area is boxed{n^3}.

🧠:[<think>Okay, so I have this problem here: two lines given by the equations ax - y - 2 = 0 and (2 - a)x - y + 1 = 0 are parallel. I need to find the value of 'a' from the options given. Let me think about how to approach this.First, I remember that for two lines to be parallel, their slopes must be equal. So, if I can find the slopes of these two lines and set them equal to each other, I can solve for 'a'. That makes sense. Let me try that.The general form of a line is often written as y = mx + b, where m is the slope. So, maybe I should convert both equations into this slope-intercept form to find their slopes.Starting with the first equation: ax - y - 2 = 0. Let me rearrange this. If I add y to both sides and add 2 to both sides, I get y = ax - 2. So, the slope (m1) of the first line is 'a'.Now, the second equation: (2 - a)x - y + 1 = 0. Let me do the same here. Add y to both sides and subtract 1 from both sides. That gives y = (2 - a)x + 1. So, the slope (m2) of the second line is (2 - a).Since the lines are parallel, their slopes must be equal. Therefore, I can set m1 = m2:a = 2 - aHmm, solving for 'a' here. Let's do that. Adding 'a' to both sides:a + a = 2Which simplifies to:2a = 2Then, dividing both sides by 2:a = 1Wait, so according to this, a equals 1. Let me check the options. The options are A:2, B:1, C:0, D:-1. So, B is 1. So, the answer should be B? Let me verify this again to make sure I didn't make a mistake.Let me plug a = 1 back into both equations and check if they are indeed parallel.First equation becomes 1*x - y - 2 = 0 → x - y - 2 = 0 → y = x - 2. Slope is 1.Second equation becomes (2 - 1)x - y + 1 = 0 → 1*x - y + 1 = 0 → y = x + 1. Slope is 1. Yep, both slopes are 1, so they are parallel. That checks out.Wait, but let me check another option just to be thorough. What if I choose a different value? Let's say a = 2 (option A). Then first equation is 2x - y - 2 = 0 → y = 2x - 2. Slope 2.Second equation becomes (2 - 2)x - y + 1 = 0 → 0x - y + 1 = 0 → y = 1. That's a horizontal line with slope 0. Not parallel. So, a=2 isn't correct.Similarly, if a=0 (option C), first equation is 0x - y -2 =0 → y = -2. Horizontal line, slope 0. Second equation is (2 -0)x - y +1=0 → 2x - y +1=0 → y=2x +1. Slope 2. Not equal. So, not parallel.If a=-1 (option D), first equation: -1x - y -2=0 → y = -x -2. Slope -1. Second equation: (2 - (-1))x - y +1=0 → 3x - y +1=0 → y=3x +1. Slope 3. Not equal. So, not parallel.Therefore, only when a=1 do both lines have the same slope, which is 1, and thus are parallel. So, the answer is B:1.Wait, just to make sure, is there another way this problem could be approached? Maybe by using the condition for parallelism in standard form equations. Let me recall. The standard form is Ax + By + C = 0. For two lines A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 to be parallel, the coefficients must satisfy A1/B1 = A2/B2, provided that B1 and B2 are not zero. Is that correct?Let me check that. If two lines are parallel, their slopes are equal. The slope for standard form is -A/B. So, slope1 = -A1/B1, slope2 = -A2/B2. So, setting them equal: -A1/B1 = -A2/B2 ⇒ A1/B1 = A2/B2. So yes, that condition holds.So, applying this to the given equations. First equation is ax - y -2 =0. Here, A1 = a, B1 = -1. Second equation is (2 - a)x - y +1 =0. Here, A2 = (2 - a), B2 = -1.So, the condition is A1/B1 = A2/B2. Plugging in:a / (-1) = (2 - a)/ (-1)Multiply both sides by (-1):a = 2 - aWhich is the same equation as before. So, 2a = 2 ⇒ a=1. Same result. So, this confirms the answer is indeed B:1.I think that's thorough. Checked two different methods, verified with substitution, and ruled out other options. Confident that the answer is B.Final Answerboxed{B}</think>To determine the value of ( a ) such that the lines ( ax - y - 2 = 0 ) and ( (2 - a)x - y + 1 = 0 ) are parallel, we need to ensure their slopes are equal.First, we convert both equations to slope-intercept form ( y = mx + b ):1. For the first line ( ax - y - 2 = 0 ): [ ax - y - 2 = 0 implies y = ax - 2 ] The slope ( m_1 ) is ( a ).2. For the second line ( (2 - a)x - y + 1 = 0 ): [ (2 - a)x - y + 1 = 0 implies y = (2 - a)x + 1 ] The slope ( m_2 ) is ( 2 - a ).Since the lines are parallel, their slopes must be equal:[a = 2 - a]Solving for ( a ):[a + a = 2 implies 2a = 2 implies a = 1]Verification by substituting ( a = 1 ) back into the equations confirms both lines have a slope of 1, making them parallel. Checking other options confirms they do not satisfy the parallelism condition.Thus, the value of ( a ) is (boxed{B}).,<think>Okay, let's see. The problem is about two lines being parallel and finding the value of 'a'. The lines are given by the equations ax - y - 2 = 0 and (2 - a)x - y + 1 = 0. The options are A: 2, B: 1, C: 0, D: -1. Hmm. I need to figure out which value of 'a' makes these two lines parallel.First, I remember that two lines are parallel if their slopes are equal. So, if I can find the slopes of both lines, set them equal, and solve for 'a', that should give me the answer. Let me recall how to find the slope of a line given in standard form. The standard form is usually Ax + By + C = 0, and the slope is -A/B. Is that right? Let me verify. Suppose we have a line like 2x - 3y + 4 = 0. Then, solving for y gives 3y = 2x + 4, so y = (2/3)x + 4/3. The slope is 2/3, which is equal to -A/B where A is 2 and B is -3. So, -2/(-3) = 2/3. Yeah, that works. So the formula for slope in standard form Ax + By + C = 0 is indeed -A/B.So, applying that to both lines. The first line is ax - y - 2 = 0. Here, A is 'a' and B is '-1', so the slope would be -a / (-1) = a. Let me check that again. If the equation is ax - y - 2 = 0, then to convert it to slope-intercept form (y = mx + b), we can rearrange:ax - y - 2 = 0 => -y = -ax + 2 => y = ax - 2.So the slope is indeed 'a', which matches what I got using the formula. Good.Now the second line is (2 - a)x - y + 1 = 0. Let's do the same. A is (2 - a), B is -1. So the slope would be -(2 - a)/(-1) = (2 - a)/1 = 2 - a. Alternatively, converting to slope-intercept form:(2 - a)x - y + 1 = 0 => -y = -(2 - a)x - 1 => y = (2 - a)x + 1.So the slope is (2 - a). Perfect, so that also matches.Since the lines are parallel, their slopes must be equal. So set the slopes equal to each other:a = 2 - a.Solving for 'a', we can add 'a' to both sides:a + a = 2 2a = 2 a = 1.So, the value of 'a' should be 1, which is option B. Let me just verify once again to make sure I didn't make a mistake.First line's slope: a, if a = 1, then slope is 1.Second line's slope: 2 - a, if a =1, then slope is 2 -1 =1. So both slopes are 1, which are equal. Therefore, the lines are parallel when a =1.Looking at the answer choices, B is 1, so that's the answer. Let me just check if any other options might also work, just in case. Let's test each option:Option A: a =2. Then first line's slope is 2, second line's slope is 2 -2 =0. So slopes 2 and 0 are not equal. Not parallel.Option B: a=1. Slopes 1 and 1. Equal. Parallel.Option C: a=0. First slope 0, second slope 2 -0 =2. Not equal.Option D: a=-1. First slope -1, second slope 2 -(-1)=3. Not equal. So only option B works. Therefore, the correct answer is B.Final Answerboxed{B}</think>To determine the value of (a) such that the lines (ax - y - 2 = 0) and ((2 - a)x - y + 1 = 0) are parallel, we need to ensure their slopes are equal.1. Find the slope of the first line (ax - y - 2 = 0): - Rewrite in slope-intercept form: (y = ax - 2). - The slope is (a).2. Find the slope of the second line ((2 - a)x - y + 1 = 0): - Rewrite in slope-intercept form: (y = (2 - a)x + 1). - The slope is (2 - a).3. Set the slopes equal for parallel lines: [ a = 2 - a ] Solving for (a): [ 2a = 2 implies a = 1 ]4. Verification: - For (a = 1), the slopes of both lines are (1), confirming they are parallel.Thus, the value of (a) is (boxed{B}).]

🔑:Since the two lines ax-y-2=0 and (2-a)x-y+1=0 are parallel, then dfrac {a}{2-a}= dfrac {-1}{-1}, solving this gives a=1. Therefore, the correct choice is boxed{B}. This problem is solved by utilizing the property of parallel lines. It tests the method of finding real values and is a basic question. When solving, it is important to carefully read the problem and properly apply the property of parallel lines.

🧠:[<think>Okay, let's see. I need to solve this system of equations for real numbers x and y, and find all possible values of the product xy. The equations are:First equation: (x)/(x²y² - 1) - 1/x = 4Second equation: (x²y)/(x²y² - 1) + y = 2Hmm, both equations have denominators involving x²y² - 1. Maybe I can simplify by letting u = x²y² - 1? Or maybe substitute variables to make it easier. Let me check.Alternatively, notice that both equations have terms divided by (x²y² - 1). Let me denote that denominator as D for simplicity, so D = x²y² - 1. Then the first equation becomes x/D - 1/x = 4, and the second equation is (x²y)/D + y = 2.Maybe if I manipulate the first equation to express 1/D in terms of x and substitute into the second equation? Let's try.Starting with the first equation:x/D - 1/x = 4Multiply both sides by D to eliminate the denominator:x - D/x = 4DWait, but D = x²y² - 1, so substituting back:x - (x²y² - 1)/x = 4(x²y² - 1)Simplify the left side: x - (x²y² - 1)/x = x - x y² + 1/xSo left side is x - x y² + 1/x, and right side is 4x²y² - 4.So equation becomes:x - x y² + 1/x = 4x² y² - 4Hmm, that looks complicated. Maybe there's a better way. Let's look at the second equation:(x²y)/D + y = 2Again, substitute D = x²y² - 1:(x²y)/(x²y² - 1) + y = 2Maybe factor out y?y[ x²/(x²y² - 1) + 1 ] = 2But I don't know if that helps. Alternatively, let's try to express both equations in terms of D and see if we can solve for x and y.From the first equation:x/D - 1/x = 4Let's write this as x/D = 4 + 1/xSimilarly, the second equation:x²y/D + y = 2Factor out y: y(x²/D + 1) = 2Let me denote A = x/D and B = x²/D. Then from the first equation: A - 1/x = 4, so A = 4 + 1/xFrom the second equation: y(B + 1) = 2But B = x²/D = x * (x/D) = x*A. Since A = 4 + 1/x, then B = x*(4 + 1/x) = 4x + 1Therefore, the second equation becomes y*(4x + 1 + 1) = y*(4x + 2) = 2So we have:y*(4x + 2) = 2Therefore, y = 2 / (4x + 2) = 2 / [2(2x + 1)] ) = 1/(2x + 1)So y = 1/(2x + 1). That's helpful. Now we can substitute this into the first equation.Wait, let's check that. So if we can express y in terms of x, then we can substitute into the first equation. So y = 1/(2x + 1). Let's substitute this into the first equation.First, the first equation was:x/D - 1/x = 4, where D = x²y² - 1Since y = 1/(2x + 1), let's compute y²:y² = 1/(2x + 1)^2So D = x² * [1/(2x + 1)^2] - 1 = x²/(2x + 1)^2 - 1Therefore, D = [x² - (2x + 1)^2 ] / (2x + 1)^2Compute numerator:x² - (4x² + 4x + 1) = x² - 4x² - 4x -1 = -3x² -4x -1So D = (-3x² -4x -1)/(2x +1)^2Now substitute D into the first equation:x/D - 1/x = 4So x/D = 4 + 1/xTherefore, D = x / (4 + 1/x) = x / [(4x +1)/x] = x^2 / (4x +1)But we also have D = (-3x² -4x -1)/(2x +1)^2Therefore, set them equal:x² / (4x +1) = (-3x² -4x -1)/(2x +1)^2Cross-multiplying:x²*(2x +1)^2 = (-3x² -4x -1)*(4x +1)Let me compute both sides.Left side: x²*(4x² +4x +1) = 4x^4 +4x^3 +x²Right side: (-3x² -4x -1)*(4x +1). Let's expand this.Multiply term by term:-3x²*4x = -12x³-3x²*1 = -3x²-4x*4x = -16x²-4x*1 = -4x-1*4x = -4x-1*1 = -1Combine terms:-12x³ -3x² -16x² -4x -4x -1Simplify:-12x³ -19x² -8x -1So left side: 4x^4 +4x^3 +x²Right side: -12x³ -19x² -8x -1Bring all terms to the left side:4x^4 +4x^3 +x² +12x³ +19x² +8x +1 =0Combine like terms:4x^4 + (4x³ +12x³) + (x² +19x²) +8x +1 =0Which is:4x^4 +16x³ +20x² +8x +1 =0So we have a quartic equation: 4x^4 +16x³ +20x² +8x +1 =0Hmm, quartic equations can be tough. Let's see if we can factor this.First, check for rational roots using Rational Root Theorem. Possible roots are ±1, ±1/2, ±1/4.Test x = -1:4*(-1)^4 +16*(-1)^3 +20*(-1)^2 +8*(-1) +1 = 4 -16 +20 -8 +1 = 1. Not zero.x = -1/2:4*(1/16) +16*(-1/8) +20*(1/4) +8*(-1/2) +1= 0.25 -2 +5 -4 +1 = 0.25 -2 is -1.75 +5 is 3.25 -4 is -0.75 +1 is 0.25. Not zero.x = -1/4:4*(1/256) +16*(-1/64) +20*(1/16) +8*(-1/4) +1= 0.015625 -0.25 +1.25 -2 +1= 0.015625 -0.25 = -0.234375 +1.25 = 1.015625 -2 = -0.984375 +1 = 0.015625. Not zero.x=1: 4 +16 +20 +8 +1=49 ≠0x=1/2: 4*(1/16) +16*(1/8) +20*(1/4) +8*(1/2)+1=0.25+2+5+4+1=12.25≠0So no rational roots. Maybe the quartic factors into quadratics?Suppose 4x^4 +16x³ +20x² +8x +1 = (ax² +bx +c)(dx² +ex +f)Multiply out:ad x^4 + (ae + bd)x³ + (af + be + cd)x² + (bf + ce)x + cfWe need ad=4, so possible a=2, d=2 or a=4,d=1, etc.Try a=2, d=2:Then:ad=4ae + bd=16: 2e +2b=16 => e +b=8af + be + cd=20: 2f +b e +2c=20bf + ce=8: b f +c e=8cf=1: c f=1Since c f=1. Assuming c and f are integers. So possible c=1, f=1 or c=-1,f=-1Let's try c=1, f=1:Then:From bf + ce =8: b*1 +1*e=8 => b + e=8. But from earlier e +b=8. So same as before.Now, from af + be + cd=20: 2*1 +b e +2*1=20 => 2 +b e +2=20 => b e=16We have e +b=8 and b e=16So solving for b and e:Let me call s = b + e =8, p = b e=16So quadratic equation: t^2 -8t +16=0. Discriminant 64 -64=0. So double root t=4. So b=4, e=4So factors are (2x² +4x +1)(2x² +4x +1). Wait, let's check.Wait, a=2, d=2, b=4, e=4, c=1, f=1.So first quadratic: 2x² +4x +1Second quadratic:2x² +4x +1Multiply: (2x² +4x +1)^2 =4x^4 +16x³ + (16x² +4x²) +8x +1 =4x^4 +16x³ +20x² +8x +1. Exactly! Perfect square.Therefore, the quartic factors as (2x² +4x +1)^2=0So the equation is (2x² +4x +1)^2=0 => 2x² +4x +1=0Solve for x:x = [-4 ±√(16 -8)] / (2*2) = [-4 ±√8]/4 = [-4 ±2√2]/4 = [-2 ±√2]/2So x = (-2 +√2)/2 or x = (-2 -√2)/2Simplify:x = -1 + (√2)/2 or x = -1 - (√2)/2Therefore, two possible x values. Now, let's find the corresponding y values using y =1/(2x +1)First, for x = (-2 +√2)/2:Compute 2x +1:2x +1 =2*(-2 +√2)/2 +1 = (-2 +√2) +1 = -1 +√2So y =1/( -1 +√2 )Multiply numerator and denominator by (1 +√2):y = (1 +√2)/[ (-1 +√2)(1 +√2) ] = (1 +√2)/[ (-1)(1) + (-1)(√2) +√2(1) +√2*√2 ]Simplify denominator:-1 -√2 +√2 +2 = (-1 +2) + (-√2 +√2) =1 +0=1Therefore, y=1 +√2Similarly, for x = (-2 -√2)/2:Compute 2x +1 =2*(-2 -√2)/2 +1 = (-2 -√2) +1 = -1 -√2Thus y =1/( -1 -√2 )Multiply numerator and denominator by (-1 +√2):y = (-1 +√2)/[ (-1 -√2)(-1 +√2) ]Denominator: (-1)^2 - (√2)^2 =1 -2= -1Therefore, y= (-1 +√2)/(-1) =1 -√2So we have two solutions:1. x = (-2 +√2)/2 and y=1 +√22. x = (-2 -√2)/2 and y=1 -√2Now, compute xy for both cases.First case:x = (-2 +√2)/2, y=1 +√2xy = [ (-2 +√2)/2 ]*(1 +√2 )Multiply numerator:(-2 +√2)(1 +√2 ) = -2*1 + (-2)*√2 +√2*1 +√2*√2= -2 -2√2 +√2 +2= (-2 +2) + (-2√2 +√2 )= 0 -√2 = -√2Divide by 2: -√2 /2Wait, wait. Wait, [ (-2 +√2)/2 ]*(1 +√2 ) is equal to (-2 +√2)(1 +√2)/2Compute numerator:(-2)(1) + (-2)(√2) + √2(1) +√2(√2)= -2 -2√2 +√2 +2= (-2 +2) + (-2√2 +√2)= 0 -√2 = -√2So numerator is -√2, divided by 2: xy = -√2 /2Second case:x = (-2 -√2)/2, y=1 -√2xy = [ (-2 -√2)/2 ]*(1 -√2 )Multiply numerator:(-2 -√2)(1 -√2 ) = -2*1 + (-2)(-√2) + (-√2)*1 + (-√2)(-√2)= -2 +2√2 -√2 +2= (-2 +2) + (2√2 -√2 )=0 +√2 =√2Divide by 2: √2 /2Therefore, the product xy is either -√2 /2 or √2 /2Wait, but wait, let me check the calculations again because I might have messed up the signs.First case:x = (-2 +√2)/2, y =1 +√2Multiply x and y:[ (-2 +√2)/2 ] * (1 +√2 )Multiply out:[ (-2)(1) + (-2)(√2) +√2(1) +√2*√2 ] /2= [ -2 -2√2 +√2 +2 ] /2= [ (-2 +2) + (-2√2 +√2 ) ] /2= [ 0 -√2 ] /2 = (-√2)/2Second case:x = (-2 -√2)/2, y=1 -√2Multiply:[ (-2 -√2)/2 ]*(1 -√2 )= [ (-2)(1) + (-2)(-√2) + (-√2)(1) + (-√2)(-√2) ] /2= [ -2 +2√2 -√2 +2 ] /2= [ (-2 +2) + (2√2 -√2 ) ] /2= [0 +√2 ] /2 = √2 /2So the products are -√2/2 and √2/2. Therefore, the possible values of xy are ±√2/2.But let me check if both solutions are valid. Since we squared terms and did substitutions, we need to verify that original equations are satisfied and denominators are not zero.First, check denominators. Original denominators are x²y² -1. Let's compute x²y² -1 for both solutions.First solution: x = (-2 +√2)/2, y=1 +√2Compute x²y²:First compute x:x = (-2 +√2)/2 ≈ (-2 +1.414)/2 ≈ (-0.586)/2 ≈ -0.293But exact value: x = (-2 +√2)/2, y=1 +√2x² = [ (-2 +√2)/2 ]^2 = (4 -4√2 +2)/4 = (6 -4√2)/4 = (3 -2√2)/2y² = (1 +√2)^2 =1 +2√2 +2 =3 +2√2x² y² = (3 -2√2)/2 * (3 +2√2) = [ (3)^2 - (2√2)^2 ] /2 = (9 -8)/2=1/2Therefore x²y² -1=1/2 -1= -1/2 ≠0, so denominator is non-zero. Okay.Similarly, second solution: x=(-2 -√2)/2, y=1 -√2x² = [ (-2 -√2)/2 ]^2 = (4 +4√2 +2)/4= (6 +4√2)/4= (3 +2√2)/2y²=(1 -√2)^2=1 -2√2 +2=3 -2√2x²y²=(3 +2√2)/2*(3 -2√2)= [9 - (2√2)^2]/2=(9 -8)/2=1/2Therefore x²y² -1=1/2 -1= -1/2 ≠0. So denominators are okay.Also check the original equations.First solution: x = (-2 +√2)/2, y=1 +√2First equation:x/(x²y² -1) -1/x = [ (-2 +√2)/2 ] / (-1/2) - 1/[ (-2 +√2)/2 ]Simplify:[ (-2 +√2)/2 ] / (-1/2 ) = [ (-2 +√2)/2 ] * (-2 )= (-2 +√2)*(-1)= 2 -√2Then subtract 1/x: 1/[ (-2 +√2)/2 ]= 2/(-2 +√2)= multiply numerator and denominator by (-2 -√2):2*(-2 -√2)/[ (-2 +√2)(-2 -√2) ]= (-4 -2√2)/[4 -2]= (-4 -2√2)/2= -2 -√2Therefore, first equation: 2 -√2 - (-2 -√2 )=2 -√2 +2 +√2=4. Which matches the first equation.Second equation: (x²y)/(x²y² -1 ) + y = [ (x²y)/(-1/2) ] + y= -2x²y + yCompute x²y:x²= [ (-2 +√2)/2 ]^2= (4 -4√2 +2)/4= (6 -4√2)/4= (3 -2√2)/2Multiply by y: (3 -2√2)/2 * (1 +√2 )= [3(1) +3√2 -2√2(1) -2√2*√2 ] /2= [3 +3√2 -2√2 -4 ] /2= [ (3-4) + (3√2 -2√2) ] /2= [ -1 +√2 ] /2So -2x²y= -2*(-1 +√2)/2= (1 -√2 )Add y: (1 -√2 ) + (1 +√2 )=2. Which matches the second equation.Similarly, check the second solution.x=(-2 -√2)/2, y=1 -√2First equation:x/(x²y² -1 ) -1/x = [ (-2 -√2)/2 ] / (-1/2 ) -1/[ (-2 -√2)/2 ]Compute first term: [ (-2 -√2)/2 ] / (-1/2 )= [ (-2 -√2)/2 ] * (-2 )= (2 +√2 )Second term: 1/[ (-2 -√2)/2 ]=2/(-2 -√2 )= multiply numerator and denominator by (-2 +√2 ):2*(-2 +√2 )/[ (-2 -√2 )(-2 +√2 ) ]= (-4 +2√2 )/(4 -2)= (-4 +2√2 )/2= -2 +√2Therefore, first equation: (2 +√2 ) - (-2 +√2 )=2 +√2 +2 -√2=4. Correct.Second equation:(x²y)/(x²y² -1 ) + y= -2x²y + yCompute x²y:x²= [ (-2 -√2)/2 ]^2= (4 +4√2 +2)/4= (6 +4√2)/4=(3 +2√2)/2Multiply by y=1 -√2:(3 +2√2)/2*(1 -√2 )= [3(1) -3√2 +2√2(1) -2√2*√2 ] /2= [3 -3√2 +2√2 -4 ] /2= [ (3 -4 ) + (-3√2 +2√2 ) ] /2= [ -1 -√2 ] /2So -2x²y= -2*(-1 -√2 )/2= (1 +√2 )Add y: (1 +√2 ) + (1 -√2 )=2. Correct.So both solutions satisfy the original equations. Therefore, the possible values of xy are -√2 /2 and √2 /2.But the problem asks for all possible values of the product xy, so they are two values: ±√2/2. However, in the exact form, √2/2 and -√2/2, which can be written as ±√2/2.But perhaps rationalizing, √2/2 is equal to 1/√2, but maybe better left as is.Alternatively, they might want the answer in simplified radical form. Both are acceptable, but √2/2 is standard.Therefore, the possible values are √2/2 and -√2/2.Final AnswerThe possible values of the product ( xy ) are (boxed{dfrac{sqrt{2}}{2}}) and (boxed{-dfrac{sqrt{2}}{2}}).</think>Given the system of equations:[left{begin{array}{l}frac{x}{x^{2} y^{2} - 1} - frac{1}{x} = 4 frac{x^{2} y}{x^{2} y^{2} - 1} + y = 2end{array}right.]we start by denoting ( D = x^2 y^2 - 1 ). This allows us to rewrite the equations as:1. (frac{x}{D} - frac{1}{x} = 4)2. (frac{x^2 y}{D} + y = 2)From the second equation, we factor out ( y ):[y left( frac{x^2}{D} + 1 right) = 2]Let ( A = frac{x}{D} ). Then, from the first equation, we have ( A - frac{1}{x} = 4 ), which gives ( A = 4 + frac{1}{x} ). Substituting ( A ) into the second equation, we get:[y left( x left( 4 + frac{1}{x} right) + 1 right) = 2 implies y (4x + 2) = 2 implies y = frac{2}{4x + 2} = frac{1}{2x + 1}]Substituting ( y = frac{1}{2x + 1} ) into the first equation, we compute ( D ):[D = x^2 left( frac{1}{2x + 1} right)^2 - 1 = frac{x^2}{(2x + 1)^2} - 1 = frac{x^2 - (2x + 1)^2}{(2x + 1)^2} = frac{-3x^2 - 4x - 1}{(2x + 1)^2}]Substituting ( D ) back into the first equation and solving, we find:[4x^4 + 16x^3 + 20x^2 + 8x + 1 = 0]This factors as:[(2x^2 + 4x + 1)^2 = 0 implies 2x^2 + 4x + 1 = 0]Solving for ( x ):[x = frac{-2 pm sqrt{2}}{2}]For each ( x ), we find ( y ):- If ( x = frac{-2 + sqrt{2}}{2} ), then ( y = 1 + sqrt{2} )- If ( x = frac{-2 - sqrt{2}}{2} ), then ( y = 1 - sqrt{2} )Calculating the product ( xy ):- For ( x = frac{-2 + sqrt{2}}{2} ) and ( y = 1 + sqrt{2} ): [ xy = left( frac{-2 + sqrt{2}}{2} right) (1 + sqrt{2}) = -frac{sqrt{2}}{2} ]- For ( x = frac{-2 - sqrt{2}}{2} ) and ( y = 1 - sqrt{2} ): [ xy = left( frac{-2 - sqrt{2}}{2} right) (1 - sqrt{2}) = frac{sqrt{2}}{2} ]Thus, the possible values of the product ( xy ) are (boxed{dfrac{sqrt{2}}{2}}) and (boxed{-dfrac{sqrt{2}}{2}}).,<think>Okay, so I have this system of equations to solve:left{begin{array}{l}frac{x}{x^{2} y^{2}-1} - frac{1}{x} = 4 frac{x^{2} y}{x^{2} y^{2}-1} + y = 2end{array}right.And I need to find all possible values of the product ( xy ). Hmm, let's start by trying to understand the structure of these equations. Both equations have denominators involving ( x^2 y^2 - 1 ), which suggests that maybe substituting ( u = x^2 y^2 - 1 ) could be helpful, but I'm not sure yet. Alternatively, maybe we can manipulate the equations to express them in terms of ( xy ), which is the product we need to find. Let me see.Looking at the first equation:frac{x}{x^{2} y^{2} - 1} - frac{1}{x} = 4Let me try to combine the terms on the left side over a common denominator. The common denominator would be ( x(x^2 y^2 - 1) ). So:frac{x cdot x - (x^2 y^2 - 1)}{x(x^2 y^2 - 1)} = 4Simplifying the numerator:( x cdot x = x^2 ), so:frac{x^2 - (x^2 y^2 - 1)}{x(x^2 y^2 - 1)} = 4Which becomes:frac{x^2 - x^2 y^2 + 1}{x(x^2 y^2 - 1)} = 4Factor out ( x^2 ) from the first two terms in the numerator:frac{x^2(1 - y^2) + 1}{x(x^2 y^2 - 1)} = 4Hmm, not sure if that helps. Let's compute the numerator and denominator more directly:Numerator: ( x^2 - x^2 y^2 + 1 = x^2(1 - y^2) + 1 )Denominator: ( x(x^2 y^2 - 1) )So maybe rewrite the denominator as ( x(x^2 y^2 - 1) = x^3 y^2 - x )But perhaps there's a smarter substitution here. Let me denote ( z = x y ). Then ( x^2 y^2 = z^2 ), so the denominator in both equations becomes ( z^2 - 1 ). Let me try that.Let ( z = xy ). Then ( x^2 y^2 = z^2 ), so the equations become:First equation:frac{x}{z^2 - 1} - frac{1}{x} = 4Second equation:frac{x^2 y}{z^2 - 1} + y = 2But in the second equation, ( x^2 y = x cdot x y = x z ), since ( z = xy ). So substitute that in:Second equation becomes:frac{x z}{z^2 - 1} + y = 2But ( y = frac{z}{x} ), since ( z = x y implies y = z / x ). Substitute that into the second equation:frac{x z}{z^2 - 1} + frac{z}{x} = 2So now, the first equation is:frac{x}{z^2 - 1} - frac{1}{x} = 4And the second equation is:frac{x z}{z^2 - 1} + frac{z}{x} = 2So now we have two equations in terms of ( x ) and ( z ). Maybe we can solve for ( x ) in terms of ( z ) from the first equation and substitute into the second equation. Let's try that.First equation:frac{x}{z^2 - 1} - frac{1}{x} = 4Multiply both sides by ( x(z^2 - 1) ) to eliminate denominators:Left side: ( x cdot x - (z^2 - 1) = x^2 - z^2 + 1 )Right side: ( 4 x(z^2 - 1) )So:x^2 - z^2 + 1 = 4 x(z^2 - 1)Let me rearrange this:x^2 - 4x(z^2 - 1) - z^2 + 1 = 0This is a quadratic equation in terms of ( x ). Let's write it as:x^2 - 4(z^2 - 1)x - (z^2 - 1) = 0Wait, hold on:Wait, expanding 4 x(z^2 -1):So the equation is:x^2 - 4(z^2 -1)x - z^2 + 1 = 0Hmm, perhaps factor out the terms. Let's see. Alternatively, maybe use quadratic formula to solve for x in terms of z. Let's try that.Let me write the equation as:x^2 - 4(z^2 -1)x + (- z^2 +1) = 0So quadratic in x:x^2 + [ -4(z^2 -1) ] x + ( - z^2 +1 ) = 0Using quadratic formula:x = [4(z^2 -1) ± sqrt(16(z^2 -1)^2 - 4 * 1 * (-z^2 +1))]/2Simplify discriminant:16(z^2 -1)^2 -4*(-z^2 +1) = 16(z^2 -1)^2 +4(z^2 -1)Factor 4(z^2 -1):= 4(z^2 -1)[4(z^2 -1) +1]= 4(z^2 -1)(4z^2 -4 +1) = 4(z^2 -1)(4z^2 -3)Therefore, x = [4(z^2 -1) ± sqrt(4(z^2 -1)(4z^2 -3))]/2Simplify sqrt term:sqrt(4(z^2 -1)(4z^2 -3)) = 2 sqrt( (z^2 -1)(4z^2 -3) )Therefore, x = [4(z^2 -1) ± 2 sqrt( (z^2 -1)(4z^2 -3) ) ] / 2Factor out 2 in numerator:x = 2(z^2 -1) ± sqrt( (z^2 -1)(4z^2 -3) )So x = 2(z^2 -1) ± sqrt{(z^2 -1)(4z^2 -3)}Hmm, this is getting complicated. Maybe there's another way. Let's look at the second equation:Second equation:frac{x z}{z^2 - 1} + frac{z}{x} = 2Let me denote the first term as A and the second term as B:A = (x z)/(z² -1)B = z / xSo A + B = 2But from the first equation, maybe we can express A or B in terms of the other.Wait, the first equation was:( x ) / (z² -1 ) - 1/x = 4Multiply both sides by z:( x z ) / (z² -1 ) - z / x = 4 zBut notice that (x z)/(z² -1) is A, and z/x is B. So this gives:A - B = 4 zBut from the second equation, we have A + B = 2So now we have a system:A - B = 4 zA + B = 2We can solve for A and B:Adding the two equations:2A = 4 z + 2 => A = 2 z + 1Subtracting the two equations:2B = 2 - 4 z => B = 1 - 2 zBut A = (x z)/(z² -1 ) = 2 z + 1And B = z / x = 1 - 2 zTherefore, from B: z / x = 1 - 2 z => x = z / (1 - 2 z )Similarly, from A: (x z ) / (z² -1 ) = 2 z + 1But we can substitute x from B into A:x = z / (1 - 2 z )Therefore:( [ z / (1 - 2 z ) ] * z ) / ( z² - 1 ) = 2 z + 1Simplify numerator:z² / (1 - 2 z )Denominator: z² -1Therefore:[ z² / (1 - 2 z ) ] / ( z² -1 ) = 2 z + 1Which is:z² / [ (1 - 2 z )( z² -1 ) ] = 2 z + 1Multiply both sides by (1 - 2 z )( z² -1 ):z² = (2 z +1 )(1 - 2 z )( z² -1 )Let me compute the right side.First, compute (2 z +1 )(1 - 2 z ):This is a difference of squares: (1)^2 - (2 z)^2 = 1 -4 z², but wait, no:Wait, (a + b)(a - b) = a² - b², but here (2z +1)(1 -2 z ) is actually -(2z -1)(2z +1) which is -(4 z² -1). Let me compute it:(2 z +1)(1 -2 z ) = 2 z *1 + 2 z * (-2 z ) +1*1 +1*(-2 z ) = 2 z -4 z² +1 -2 z = (2 z -2 z ) + (-4 z² ) +1 = -4 z² +1So indeed, (2 z +1)(1 -2 z ) = 1 -4 z²Therefore, the right side is (1 -4 z²)( z² -1 )So the equation becomes:z² = (1 -4 z²)( z² -1 )Expand the right side:(1)(z² -1 ) -4 z² ( z² -1 ) = z² -1 -4 z^4 +4 z² = (z² +4 z² ) + (-1 ) + (-4 z^4 ) = 5 z² -1 -4 z^4Therefore, the equation is:z² = -4 z^4 +5 z² -1Bring all terms to left side:z² +4 z^4 -5 z² +1 =0Combine like terms:4 z^4 -4 z² +1 =0So quartic equation: 4 z^4 -4 z² +1 =0Let me set w = z², so the equation becomes:4 w² -4 w +1 =0Solve for w:w = [4 ± sqrt(16 -16 ) ] /8 = [4 ±0 ] /8 = 4/8 = 1/2Thus, w=1/2, so z²=1/2 => z=±√(1/2)=±(√2)/2Thus, z=±√2/2Therefore, the possible values of z=xy are ±√2/2. However, we need to check if these solutions satisfy the original equations and if there are any restrictions.Wait, but let's check if these solutions are valid. Because when we squared equations or did manipulations, sometimes extraneous solutions can be introduced.So first, let's check z=√2/2 and z=-√2/2.First, check if these z values lead to valid x and y.From earlier, we have x = z / (1 - 2 z )So let's compute x for each z:Case 1: z=√2/2Compute 1 -2 z =1 -2*(√2/2)=1 -√2Therefore, x= (√2/2)/(1 -√2 )Multiply numerator and denominator by (1 +√2 ) to rationalize:x= (√2/2)(1 +√2 ) / [ (1 -√2)(1 +√2 ) ] = (√2/2)(1 +√2 ) / (1 -2 )= (√2/2)(1 +√2 ) / (-1 )= - (√2/2)(1 +√2 )Similarly, y= z /x = (√2/2 ) / [ - (√2/2)(1 +√2 ) ] = -1/(1 +√2 )Multiply numerator and denominator by (1 -√2 ):- (1 -√2 ) / [ (1 +√2 )(1 -√2 ) ] = - (1 -√2 ) / (1 -2 ) = - (1 -√2 ) / (-1 ) = (1 -√2 )Wait, so y=(1 -√2 )Now, check if the original denominators are non-zero. The denominator in original equations is x² y² -1.Compute x² y²:x = - (√2/2)(1 +√2 )x² = ( (√2/2)^2 )(1 +√2 )² = ( (2/4 ) )(1 +2√2 +2 ) = (1/2)(3 +2√2 )y =1 -√2y² = (1 -√2 )² =1 -2√2 +2=3 -2√2Therefore, x² y²=(1/2)(3 +2√2 )(3 -2√2 )Multiply (3 +2√2 )(3 -2√2 )=9 - (2√2 )²=9 -8=1So x² y²=(1/2)(1 )=1/2Therefore, denominator x² y² -1=1/2 -1=-1/2 ≠0, so denominators are okay.Similarly, check the other case.Case 2: z=-√2/2Compute x= z/(1 -2 z )= (-√2/2 ) / [1 -2*(-√2/2 ) ]= (-√2/2 ) / (1 +√2 )Rationalize denominator:x= (-√2/2 )(1 -√2 ) / [ (1 +√2 )(1 -√2 ) ] = (-√2/2 )(1 -√2 ) / (1 -2 )= (-√2/2 )(1 -√2 ) / (-1 )= (√2/2 )(1 -√2 )Then y= z /x= (-√2/2 ) / [ (√2/2 )(1 -√2 ) ] = -1/(1 -√2 )Multiply numerator and denominator by (1 +√2 ):- (1 +√2 ) / [ (1 -√2 )(1 +√2 ) ] = - (1 +√2 ) / (-1 )= (1 +√2 )Compute x² y²:x= (√2/2 )(1 -√2 )x²=( (√2/2 )² )(1 -√2 )²=(2/4 )(1 -2√2 +2 )=(1/2)(3 -2√2 )y=1 +√2y²=(1 +√2 )²=1 +2√2 +2=3 +2√2Therefore, x² y²=(1/2)(3 -2√2 )(3 +2√2 )=(1/2)(9 -8 )=(1/2)(1 )=1/2So denominator x² y² -1=1/2 -1=-1/2 ≠0, so denominators are okay here as well.Therefore, both z=√2/2 and z=-√2/2 lead to valid x and y with denominators non-zero. Therefore, both solutions are valid.But wait, the problem says "find all possible values of the product xy". Since z=xy, and we found z=±√2/2, then the possible values are √2/2 and -√2/2. Therefore, the product xy can be ±√2/2.But let me confirm by plugging back into original equations.First, take z=√2/2. Then xy=√2/2. Let's check first equation:First equation: (x)/(x² y² -1 ) -1/x =4We know x² y²=1/2, so denominator x² y² -1= -1/2So first term: x / (-1/2 )= -2 xSecond term: -1/xSo equation: -2x -1/x=4From earlier, we have x= - (√2/2)(1 +√2 )Compute -2x: -2*(- (√2/2)(1 +√2 ))=√2(1 +√2 )Compute -1/x: -1/[ - (√2/2)(1 +√2 ) ]=2/(√2(1 +√2 ))=√2/(1 +√2 )So total equation:√2(1 +√2 ) + √2/(1 +√2 ) =4?Compute first term: √2(1 +√2 )=√2 +2Second term: √2/(1 +√2 ). Multiply numerator and denominator by (1 -√2 ):√2(1 -√2 ) / (1 -2 )= √2(1 -√2 ) / (-1 )= -√2(1 -√2 )= -√2 +2Therefore, total equation:(√2 +2 ) + (-√2 +2 )=4Which is (√2 -√2 ) + (2 +2 )=0 +4=4. Correct.Similarly, check second equation:(x² y )/(x² y² -1 ) + y=2From x² y²=1/2, so denominator is -1/2First term: x² y / (-1/2 )= -2 x² ySecond term: ySo equation: -2 x² y + y=2Factor y: y(-2 x² +1 )=2Compute y=1 -√2x²=(1/2)(3 +2√2 )So x²=(3 +2√2 )/2Then -2 x² +1= -2*(3 +2√2 )/2 +1= -(3 +2√2 ) +1= -2 -2√2Therefore, y*(-2 -2√2 )= (1 -√2 )*(-2 -2√2 )Multiply:1*(-2 ) +1*(-2√2 ) -√2*(-2 ) -√2*(-2√2 )= -2 -2√2 +2√2 +2*(√2 )²= -2 -2√2 +2√2 +2*2= -2 +4=2Thus, left side is 2, right side is 2. Correct.Similarly, for z=-√2/2, similar steps would follow, but let me check one equation.Take z=-√2/2, then xy=-√2/2.First equation:(x)/(x² y² -1 ) -1/x=4Again, x² y²=1/2, denominator=-1/2First term: x/(-1/2 )=-2xSecond term:-1/xEquation: -2x -1/x=4From earlier, x= (√2/2 )(1 -√2 )Compute -2x: -2*(√2/2 )(1 -√2 )=-√2(1 -√2 )=-√2 +2Compute -1/x: -1/[ (√2/2 )(1 -√2 ) ]= -2/(√2(1 -√2 ))= -√2/(1 -√2 )Multiply numerator and denominator by (1 +√2 ):-√2(1 +√2 )/(1 -2 )= -√2(1 +√2 )/(-1 )=√2(1 +√2 )So equation:(-√2 +2 ) +√2(1 +√2 )= (-√2 +2 ) +√2 +2= (-√2 +√2 ) + (2 +2 )=0 +4=4. Correct.Similarly, check second equation:(x² y )/(x² y² -1 ) + y=2Again, denominator=-1/2, first term=-2x² ySecond term=yEquation: -2x² y + y=2Compute y=1 +√2x²=(3 -2√2 )/2Then -2x² y + y= -2*(3 -2√2 )/2*(1 +√2 ) + (1 +√2 )Simplify:- (3 -2√2 )(1 +√2 ) + (1 +√2 )Multiply (3 -2√2 )(1 +√2 ):3*1 +3*√2 -2√2*1 -2√2*√2 =3 +3√2 -2√2 -4= (3 -4 ) + (3√2 -2√2 )= -1 +√2Therefore:- ( -1 +√2 ) + (1 +√2 )=1 -√2 +1 +√2=2. Correct.Thus, both solutions are valid. Therefore, the possible values of xy are √2/2 and -√2/2.But wait, in the quartic equation, we got z²=1/2, so z=±√(1/2)=±√2/2. So these are the only solutions. Therefore, the product xy can be either √2/2 or -√2/2.Therefore, all possible values of xy are √2/2 and -√2/2.Final AnswerThe possible values of the product ( xy ) are (boxed{dfrac{sqrt{2}}{2}}) and (boxed{-dfrac{sqrt{2}}{2}}).</think>Given the system of equations:left{begin{array}{l}frac{x}{x^{2} y^{2} - 1} - frac{1}{x} = 4 frac{x^{2} y}{x^{2} y^{2} - 1} + y = 2end{array}right.we need to find all possible values of the product ( xy ).First, we introduce the substitution ( z = xy ). This transforms the equations into:1. (frac{x}{z^2 - 1} - frac{1}{x} = 4)2. (frac{xz}{z^2 - 1} + frac{z}{x} = 2)We solve these equations by expressing ( x ) in terms of ( z ). From the first equation, we find:[frac{x}{z^2 - 1} - frac{1}{x} = 4]Multiplying through by ( x(z^2 - 1) ) and simplifying, we get a quadratic equation in ( x ):[x^2 - 4(z^2 - 1)x - (z^2 - 1) = 0]Using the quadratic formula, we solve for ( x ):[x = 2(z^2 - 1) pm sqrt{(z^2 - 1)(4z^2 - 3)}]Next, we use the second equation to express ( x ) in terms of ( z ):[frac{xz}{z^2 - 1} + frac{z}{x} = 2]We find that ( x = frac{z}{1 - 2z} ) and substitute this back into the equation. After simplifying, we derive a quartic equation in ( z ):[4z^4 - 4z^2 + 1 = 0]Solving this quartic equation by setting ( w = z^2 ), we find:[w = frac{1}{2} implies z = pm frac{sqrt{2}}{2}]We verify that both ( z = frac{sqrt{2}}{2} ) and ( z = -frac{sqrt{2}}{2} ) satisfy the original equations, confirming that the product ( xy ) can indeed take these values.Thus, the possible values of the product ( xy ) are:[boxed{dfrac{sqrt{2}}{2}} text{ and } boxed{-dfrac{sqrt{2}}{2}}]]

🔑:1. The given system can be expressed as follows:left{begin{array} { l } { frac { x } { x ^ { 2 } y ^ { 2 } - 1 } - frac { 1 } { x } = 4 } { frac { x ^ { 2 } y } { x ^ { 2 } y ^ { 2 } - 1 } + y = 2 }end{array} Rightarrow left{begin{array}{l}frac{x^{2}}{x^{2} y^{2}-1}-1=4 x frac{x^{2}}{x^{2} y^{2}-1}+1=frac{2}{y}end{array}right.right.We then have4 x+frac{2}{y}=frac{2 x^{2}}{x^{2} y^{2}-1} Rightarrow 2 x+frac{1}{y}=frac{x^{2}}{x^{2} y^{2}-1}and4 x-frac{2}{y}=-2 Rightarrow 2 x-frac{1}{y}=-1 text {, }which gives usbegin{aligned}left(2 x+frac{1}{y}right)left(2 x-frac{1}{y}right) & =frac{-x^{2}}{x^{2} y^{2}-1} 4 x^{2}-frac{1}{y^{2}} & =frac{-x^{2}}{x^{2} y^{2}-1} frac{4 x^{2} y^{2}-1}{y^{2}} & =frac{-x^{2}}{x^{2} y^{2}-1} left(4 x^{2} y^{2}-1right)left(x^{2} y^{2}-1right) & =-x^{2} y^{2} 4 x^{4} y^{4}-5 x^{2} y^{2}+1 & =-x^{2} y^{2} 4 x^{4} y^{4}-4 x^{2} y^{2}+1 & =0 left(2 x^{2} y^{2}-1right)^{2} & =0 Rightarrow x^{2} y^{2}=frac{1}{2} Rightarrow x y= pm frac{1}{sqrt{2}}end{aligned}

🧠:[<think>Okay, let me try to figure out this problem. So, we have the set A = {0,1,2,...,9}, which has 10 elements. Then there are subsets B₁, B₂, ..., B_k, each of which is non-empty. The condition is that for any two different subsets B_i and B_j, the size of their intersection is at most 2. We need to find the maximum possible value of k, which is the number of such subsets.Hmm, so the problem is asking for the largest family of subsets where any two subsets share at most two elements. This seems related to combinatorics and design theory. Maybe something like a block design where intersections are controlled. Let me recall some concepts.In combinatorics, a projective plane or a balanced incomplete block design (BIBD) has specific intersection properties. But I'm not sure if that's directly applicable here. Let's think more generally.Each subset B_i can be of any size, right? The problem doesn't restrict the size of the subsets, only that they are non-empty and that any two intersect in at most two elements. So, to maximize the number of subsets, we need to arrange them such that they overlap as little as possible. But how?If there were no intersection constraints, the maximum number of subsets would just be 2¹⁰ - 1 = 1023, since each element can be either in or out. But with the intersection constraint, this number is going to be much smaller.Maybe I can approach this problem using the inclusion-exclusion principle or some combinatorial bounds. Let me try an upper bound approach.Suppose each subset B_i has size t. Then, the number of pairs in each subset is C(t,2). If any two subsets share at most two elements, then the number of pairs they share is at most C(2,2) = 1. Wait, no. If two subsets share two elements, the number of common pairs would be C(2,2) = 1. But if they share three elements, the number of common pairs would be C(3,2)=3. But in our problem, the intersection size is at most 2, so the number of common elements is ≤2, hence the number of common pairs is ≤1. Wait, maybe that's a different approach.Alternatively, think of the problem in terms of pairs. Each subset B_i contains several pairs of elements. If two subsets share more than two elements, then they share at least three elements, which would mean their intersection is at least three. But our condition is that the intersection is at most two. So, perhaps this is equivalent to saying that no two subsets share more than two elements. Therefore, the family of subsets is a 2-intersecting family? Wait, no, actually, in intersection terminology, a 2-intersecting family would mean that every pair of subsets intersects in at least two elements. But here, it's the opposite: every pair intersects in at most two elements. So, maybe this is called a family of subsets with bounded pairwise intersections.Yes, this is similar to the concept of a family of sets with bounded pairwise intersections. There's a theorem called Fisher's inequality or maybe the Erdős–Ko–Rado theorem, but Erdős–Ko–Rado typically deals with intersecting families. Wait, but in our case, it's the opposite; we want non-intersecting too much.Alternatively, think of it in terms of coding theory. Each subset can be represented as a characteristic vector in a 10-dimensional vector space over GF(2). The inner product of two vectors would be the size of the intersection modulo 2. But not sure if that's helpful here.Alternatively, use the probabilistic method or combinatorial bounds. Let's try using the Fisher-type inequality.Suppose we have k subsets, each of size t. Then, each pair of subsets shares at most two elements. Let's consider the total number of incidences between elements and subsets. If each element is in r subsets, then the total number of incidences is 10r. On the other hand, each subset has t elements, so the total number of incidences is kt. Therefore, 10r = kt. So, r = kt / 10.Now, consider the number of pairs of subsets. Each element is in r subsets, so the number of pairs of subsets that contain this element is C(r,2). Therefore, the total number of pairs of subsets that share at least one element is Σ_{x ∈ A} C(r_x,2), where r_x is the number of subsets containing x. If all r_x are equal, this would be 10 * C(r,2). But since each pair of subsets can share at most two elements, the total number of shared element pairs is at most 2 * C(k,2). Wait, not sure.Wait, each pair of subsets can share at most two elements. So, for each pair of subsets, the number of common elements is ≤2. Therefore, the total number of element pairs (x, {B_i,B_j}) where x is in both B_i and B_j is equal to Σ_{i < j} |B_i ∩ B_j| ≤ Σ_{i < j} 2 = 2 * C(k,2).But also, this total can be calculated by considering each element x, and for each x, the number of pairs of subsets that contain x is C(r_x,2). Therefore, Σ_{x ∈ A} C(r_x,2) ≤ 2 * C(k,2).So, we have the inequality:Σ_{x=0}^9 C(r_x,2) ≤ 2 * C(k,2)Using convexity, we can apply Jensen's inequality. The function C(r,2) is convex in r, so the sum is minimized when all r_x are equal or as equal as possible. Wait, but we want to bound the sum from above. Wait, since C(r,2) is convex, the sum is maximized when the variables are as unequal as possible. But we need an upper bound, so perhaps the minimal sum occurs when all r_x are equal.Wait, actually, since we need Σ C(r_x,2) ≤ 2 * C(k,2), and we want to maximize k, perhaps we can use this inequality to bound k.Assuming that each element is in r subsets, so r_x = r for all x. Then:10 * C(r,2) ≤ 2 * C(k,2)So,10 * [r(r - 1)/2] ≤ 2 * [k(k - 1)/2]Simplify:10 * r(r - 1)/2 ≤ k(k - 1)Multiply both sides by 2:10r(r - 1) ≤ k(k - 1)Also, from earlier, we have kt = 10r. So, r = kt / 10. Substitute into the inequality:10 * (kt /10)(kt /10 - 1) ≤ k(k - 1)Simplify:(kt /10)(kt /10 - 1) ≤ (k(k - 1))/10Multiply both sides by 10:kt(kt /10 - 1) ≤ k(k - 1)Divide both sides by k (assuming k ≠ 0, which it isn't):t(kt /10 - 1) ≤ (k - 1)Expand left side:(kt²)/10 - t ≤ k - 1Multiply all terms by 10:kt² - 10t ≤ 10k - 10Bring all terms to left:kt² - 10t -10k +10 ≤0Hmm, this seems messy. Maybe there's a better approach.Alternatively, consider specific subset sizes. If all subsets are pairs (size 2), then each pair can intersect with another pair in at most one element. Wait, but the problem allows intersections up to size 2. Wait, if all subsets are size 2, then two subsets can share at most one element, which is within the allowed intersection size of 2. Wait, but the problem says intersection size at most 2. So subsets of size 2 would have intersections of size at most 1. But maybe we can have larger subsets.Wait, maybe the maximum k is achieved when all subsets are pairs (size 2). Then, the maximum number of pairs is C(10,2)=45. But since each pair can intersect with another pair in at most one element, which is ≤2, so that's okay. But wait, actually, if all subsets are pairs, then any two subsets intersect in at most one element, which satisfies the condition. So, can we have 45 subsets? But the problem allows subsets of any size. If we use larger subsets, maybe we can have more subsets?Wait, but if we have larger subsets, say size 3, then two subsets of size 3 can intersect in up to 3 elements. But the problem restricts the intersection size to at most 2, so subsets of size 3 must be arranged such that no two share all three elements. But actually, even if they share two elements, that's allowed. But perhaps using larger subsets complicates things because we have to ensure that their intersections are controlled.Alternatively, maybe using subsets of varying sizes. But it's complicated.Wait, let's think about projective planes. In a projective plane of order n, each pair of lines intersects in exactly one point, and each line contains n+1 points, with the total number of lines being n² + n +1. But our case is different; we have a 10-element set, and intersections of at most 2. Maybe not directly applicable.Alternatively, think of the problem as a graph problem. Each subset is a vertex, and connect two subsets with an edge if their intersection is greater than 2. Then, our problem is to find the maximum independent set in this graph. However, this graph is complicated, and the maximum independent set is hard to compute, so maybe not helpful.Alternatively, think of the problem in terms of the inclusion matrix. But not sure.Wait, going back to the earlier inequality. If all subsets have the same size t, then we had:10 * C(r,2) ≤ 2 * C(k,2)But we also have kt =10r. Let me try plugging in t=5. Then r =k*5/10= k/2.Then:10 * C(k/2,2) ≤ 2*C(k,2)Compute:Left side: 10*( (k/2)*(k/2 -1)/2 )=10*(k(k - 2)/8)= (10k(k -2))/8= (5k(k -2))/4Right side: 2*(k(k -1)/2)=k(k -1)So inequality:(5k(k -2))/4 ≤k(k -1)Multiply both sides by 4:5k(k -2) ≤4k(k -1)Divide both sides by k (assuming k>0):5(k -2) ≤4(k -1)5k -10 ≤4k -45k -4k ≤ -4 +10k ≤6So if t=5, the maximum k would be 6. Hmm, interesting. But does such a system exist? Maybe a block design where each element is in r= k*5 /10= 3 subsets (since k=6, r=3). So each element is in 3 subsets, each subset has 5 elements, and any two subsets share at most 2 elements. Let me check if that's possible.Wait, for example, in a Steiner system S(t,k,v), but not sure. Alternatively, maybe using finite geometry. Alternatively, maybe construct such subsets.Alternatively, if t=5, k=6, each element in 3 subsets. Let's see: Each element is in 3 subsets, each subset has 5 elements, so total number of element occurrences is 6*5=30, which is 10*3=30. So that checks out.Now, the question is whether such a system can exist where any two subsets share at most 2 elements. Let me try constructing it.Take the set A={0,1,2,3,4,5,6,7,8,9}. Let's try to create 6 subsets, each of size 5, such that each element is in exactly 3 subsets, and any two subsets share at most 2 elements.This seems similar to a combinatorial design called a pairwise balanced design, but with additional regularity.Alternatively, perhaps it's a 2-design. A 2-(10,5,λ) design would have each pair of elements appearing in exactly λ blocks. But in our case, we don't want pairs to be overrepresented, but rather to limit the intersection of blocks. Wait, no, in a 2-design, the intersection of two blocks is exactly λ, but here we want the intersection to be at most 2. So perhaps not a 2-design.Alternatively, maybe using mutually orthogonal Latin squares, but I don't see the connection.Alternatively, think about the complement. If each subset is of size 5, then the complement is also size 5. Maybe considering the subsets and their complements, but not sure.Alternatively, use the concept of a code with certain distance. If each subset is a codeword, and the distance is the symmetric difference. But the intersection size relates to the dot product in some way.Alternatively, here's a concrete approach: Let me try to construct 6 subsets manually.Suppose we divide the 10 elements into two groups of 5, say Group1: 0-4 and Group2:5-9.Now, create subsets by taking combinations from both groups. For example:B1: 0,1,2,5,6B2: 0,1,3,7,8B3: 0,2,4,5,9B4: 1,3,4,6,7B5: 2,3,4,8,9B6: 5,6,7,8,9Wait, let's check intersections.B1 and B2 share 0,1 – okay, size 2.B1 and B3 share 0,2,5 – size 3. Oh, that's bad. So this doesn't work.Hmm, maybe this approach is flawed.Alternatively, use finite projective planes. Wait, a projective plane of order 3 has 13 points, which is more than 10, so not helpful.Alternatively, use the concept of a block design with block size 5 and λ=1. But a 2-(10,5,1) design would have parameters that might not exist. The number of blocks in a 2-(10,5,1) design would be C(10,2)/C(5,2) = 45 /10=4.5, which is not an integer, so such a design doesn't exist.Alternatively, use pairwise disjoint subsets. If all subsets are pairwise disjoint, then the maximum number is 10 (each subset a singleton). But we can have more subsets if we allow some overlaps, as long as overlaps are ≤2. But even two singletons would intersect in 0 elements, which is allowed, but singleton subsets can be many. Wait, but if we have all possible subsets with size ≤2, then intersections between size 2 subsets would be at most 1, which is allowed. But the problem states that subsets are non-empty, so including all subsets of size 1 and 2. The number would be C(10,1)+C(10,2)=10+45=55. But wait, but two subsets of size 2 can intersect in at most 1 element, which is okay. However, the problem allows intersections up to 2. So even if we have subsets larger than size 2, their intersections can be up to 2. So maybe including larger subsets as well?But if we include subsets of size 3, we have to ensure that any two subsets of size 3 intersect in at most 2 elements, which is possible, but if we have a subset of size 3 and a subset of size 4, they can also intersect in up to 3 elements, which is not allowed. Therefore, to include larger subsets, we need to carefully construct them so that they don't overlap too much with others.But perhaps the maximum k is achieved by taking all subsets of size ≤2. That gives 55 subsets. But perhaps we can do better by including some larger subsets as well, provided they don't intersect too much with others.Alternatively, use the concept of a hypergraph with edges of varying sizes, but with maximum pairwise intersection 2. Then, the maximum number of edges is desired.But I need to find a way to calculate or bound this.Another approach: use the inclusion matrices or linear algebra methods. Let me think.Suppose we represent each subset as a vector in a vector space over the real numbers, where each coordinate corresponds to an element of A. The inner product of two vectors would be the size of their intersection. Our condition is that the inner product of any two distinct vectors is at most 2. Then, the problem is to find the maximum number of vectors (subsets) with pairwise inner product ≤2.This resembles a spherical code problem, where vectors are points on a sphere with certain angular distances. But maybe not exactly applicable here.Alternatively, use the Johnson bound or the Fisher's inequality. Wait, in coding theory, the maximum number of codewords with certain distance. Maybe the analogous concept here is the maximum number of codewords (subsets) with pairwise intersection ≤2. Maybe there are known bounds for this.Alternatively, upper bound the number using double counting or eigenvalue methods.Wait, let's consider all possible subsets. The total number is 2¹⁰ -1=1023. But with the intersection condition, the number is much smaller. The problem is similar to a family of sets where the intersection of any two is bounded. A classical result for such families is the theorem by Fisher, Bose, and others.Wait, the theorem states that if a family of subsets has the property that the intersection of any two distinct subsets is exactly λ, then the family is a block design. But here, we have the intersection at most 2, so it's a more general case.There is a result called the Ray-Chaudhuri–Wilson theorem which gives bounds for families of subsets with intersections restricted. Let me recall.The Ray-Chaudhuri–Wilson theorem states that if we have a family of k subsets, each of size t, from a universe of size n, and the intersection of any two distinct subsets is at most s, then k ≤ C(n, s+1). But wait, in our case, the subsets can be of any size, not fixed size. So maybe a different bound.Wait, actually, the theorem in more detail: For a family of k subsets, each of size t, from an n-element set, with pairwise intersections of size at most s, then k ≤ C(n, s+1). But in our problem, subsets can be of any size, so perhaps this theorem isn't directly applicable.But maybe the theorem can be adapted. For example, if we fix the maximum subset size, then apply the theorem, but since our subsets can be of any size, maybe we need another approach.Alternatively, use the probabilistic method to find an upper bound.Alternatively, think of the problem as a graph where each vertex is a subset, and edges connect subsets that intersect in more than 2 elements. Then, the problem reduces to finding the maximum independent set in this graph. However, computing the maximum independent set is hard, but maybe we can bound its size.Alternatively, use the Lovász local lemma or other combinatorial techniques.Alternatively, think about the following: Each subset can contain at most C(10,0) + C(10,1) + C(10,2) elements? No, that doesn't make sense.Wait, perhaps another approach. For each element in A, it can be in multiple subsets, but we need to ensure that any two subsets don't share too many elements.Suppose we take all subsets of size at most 2. As before, that gives 55 subsets. Now, can we add some subsets of size 3 without violating the intersection condition?For example, take a subset of size 3, say {0,1,2}. Now, any subset of size at most 2 that intersects with this in more than 2 elements would have to be a subset of size 2 contained within {0,1,2}. But since all subsets of size 2 are already included, adding {0,1,2} would create intersections of size 2 with the subsets {0,1}, {0,2}, {1,2}, which is allowed. Wait, but {0,1,2} intersects with {0,1,2} itself? No, because we are considering distinct subsets. So, intersections with subsets of size 2 would be size 2, which is allowed. However, if we add another subset of size 3, say {0,1,3}, then the intersection with {0,1,2} is {0,1}, which is size 2, allowed. Similarly, adding {0,2,3} would intersect {0,1,2} in {0,2}, size 2. So, actually, we can add multiple subsets of size 3, as long as any two of them intersect in at most 2 elements. Wait, but two subsets of size 3 can intersect in 3 elements, which would violate the condition.So, if we add subsets of size 3, we have to make sure that any two of them intersect in at most 2 elements. So, for example, if we take all 3-element subsets that contain a fixed pair, say {0,1}, then any two of these subsets would intersect in {0,1} plus another element, so their intersection would be size 2. Wait, no. If we have two 3-element subsets both containing {0,1}, say {0,1,2} and {0,1,3}, their intersection is {0,1}, size 2, which is okay. So, we can create a family of 3-element subsets each containing a fixed pair, and the number of such subsets is 8 (since the third element can be any of the remaining 8 elements). However, each of these subsets would intersect the existing 2-element subsets {0,1}, {0,2}, {1,2}, etc., in 2 elements, which is allowed.But how many 3-element subsets can we add without overlapping too much with each other? For example, if we fix the pair {0,1}, we can have 8 subsets {0,1,x} for x in 2-9. Each of these subsets intersects with each other in {0,1}, which is size 2. So, that's allowed. So, adding 8 subsets. But also, if we fix another pair, say {0,2}, and create subsets {0,2,x}, then each of these would intersect the previous subsets {0,1,x} in {0}, which is size 1, which is okay. Similarly, if we fix multiple pairs, each time creating 8 subsets for each pair.But wait, each pair can be used to generate 8 subsets. However, if we have multiple pairs, we need to ensure that the subsets generated from different pairs don't intersect in more than 2 elements.For example, subsets from pair {0,1} are {0,1,2}, {0,1,3}, ..., {0,1,9}. Subsets from pair {0,2} are {0,2,3}, {0,2,4}, ..., {0,2,9}. Now, a subset from the first group, say {0,1,2}, and a subset from the second group, say {0,2,3}, intersect in {0,2}, which is size 2, allowed. Similarly, {0,1,3} and {0,2,3} intersect in {0,3}, size 2. Wait, no: {0,1,3} and {0,2,3} intersect in {0,3}, which is size 2. Wait, {0,1,3} and {0,2,4} would intersect in {0}, size 1. So, as long as the pairs used to generate the 3-element subsets are different, the intersections between subsets from different pair groups would be either size 1 or 2, which is okay.Therefore, perhaps we can generate multiple 3-element subsets by fixing different pairs. For each pair, we can have 8 subsets (adding each of the remaining 8 elements). However, how many pairs can we fix without overlapping too much?Wait, the total number of pairs in A is C(10,2)=45. For each pair, we can generate 8 subsets. However, each subset of size 3 contains 3 different pairs. For example, {0,1,2} contains pairs {0,1}, {0,2}, {1,2}. So, if we fix a pair {0,1} and generate subsets {0,1,x} for x=2,...,9, each of these subsets contains the pair {0,1} and another pair {0,x} or {1,x}. Therefore, if we use different pairs to generate subsets, we have to ensure that we don't reuse the same pair in multiple generators, otherwise the subsets would overlap in two elements.Wait, this is getting complicated. Maybe another approach.Suppose we use the 2-element subsets as part of our family, and then for each 2-element subset, we can extend it to a 3-element subset in 8 ways. However, if we do this for all 45 pairs, we would get 45*8=360 subsets of size 3, which is way too many and probably violating the intersection condition. Because two different 3-element subsets constructed from different pairs might intersect in 2 elements. For example, {0,1,2} (from pair {0,1}) and {0,1,3} (from pair {0,1}) intersect in {0,1}, size 2. That's allowed. But if {0,1,2} (from pair {0,1}) and {0,2,3} (from pair {0,2}) intersect in {0,2}, size 2, which is allowed. So, actually, even if we take all possible 3-element subsets, each pair of them intersect in at most 2 elements, so they satisfy the condition. Wait, but two 3-element subsets can intersect in 3 elements, which would be a problem. For example, {0,1,2} and {0,1,2} are the same subset, but since they are distinct, we have to consider two different subsets. If two different 3-element subsets share all three elements, they are actually the same subset. Wait, but in reality, different 3-element subsets can share 0,1,2, or 3 elements. But if they share 3 elements, they are the same subset, which is not allowed since the subsets must be distinct. So, the maximum intersection between two distinct subsets of size 3 is 2. Therefore, all 3-element subsets satisfy the intersection condition. Wait, that's a key point!Wait, if all subsets are of size 3, then any two distinct subsets can share at most 2 elements. Therefore, the family of all 3-element subsets of A satisfies the condition that any two subsets intersect in at most 2 elements. The number of 3-element subsets is C(10,3)=120. Similarly, if we include all subsets of size 1, 2, and 3, then the total number is C(10,1)+C(10,2)+C(10,3)=10+45+120=175. But wait, subsets of different sizes can also intersect in more than 2 elements. For example, a subset of size 3 and a subset of size 4 can intersect in 3 elements, which violates the condition. Therefore, if we include subsets of different sizes, we need to ensure that any two subsets, regardless of their sizes, intersect in at most 2 elements.Therefore, if we want to include subsets of different sizes, we have to be careful. For instance, a subset of size 4 can't share 3 elements with a subset of size 3. So, including larger subsets complicates things because they might overlap too much with smaller subsets.Therefore, maybe the maximum family is achieved by taking all subsets of size at most 2 (55 subsets) plus all subsets of size 3 (120 subsets). But as mentioned, some subsets of size 3 might intersect with subsets of size 4 or higher in more than 2 elements. Wait, no, if we don't include any subsets of size 4 or higher, then the only intersections are between subsets of size 1, 2, and 3. A subset of size 3 and a subset of size 2 can intersect in at most 2 elements (since the subset of size 2 is entirely contained within the subset of size 3 or not). Similarly, a subset of size 3 and a subset of size 1 can intersect in at most 1 element. Therefore, if we take all subsets of size ≤3, then any two subsets will intersect in at most 2 elements. Wait, is that true?Wait, take two subsets of size 3: their intersection is at most 2 elements, as they are distinct. A subset of size 3 and a subset of size 2: their intersection is at most 2 elements. A subset of size 3 and a subset of size 1: their intersection is at most 1 element. Subsets of size 2: intersections are at most 1 element. Subsets of size 1: intersections are 0 or 1 element. So yes, if we take all subsets of size ≤3, then any two distinct subsets will intersect in at most 2 elements. Therefore, this family of subsets satisfies the condition, and the number of subsets is 10 + 45 + 120 = 175.But wait, this seems too good. The problem is asking for non-empty subsets with pairwise intersections at most 2. If all subsets of size ≤3 are allowed, then 175 is the answer. But maybe I'm missing something.Wait, let's verify with an example. Suppose we have a subset of size 3: {0,1,2}, and another subset of size 4: {0,1,2,3}. Their intersection is {0,1,2}, size 3, which violates the condition. Therefore, if we include subsets of size 4, they can't coexist with subsets of size 3 that they overlap with in 3 elements. Therefore, to include subsets of size 4, we have to ensure they don't share 3 elements with any other subset.But in the previous calculation, I assumed we include all subsets of size ≤3. If we do that, but then also include some subsets of size 4, we have to make sure those subsets of size 4 don't overlap with any subsets of size 3 in 3 elements. But since there are many subsets of size 3, it's impossible to include any subset of size 4 without overlapping some subset of size 3 in 3 elements. Therefore, if we include all subsets of size ≤3, we cannot include any subsets of size ≥4. Therefore, the total number is 175.But the problem doesn't restrict subsets to size ≤3. Wait, but if we include any subset of size ≥4, we have to ensure that it doesn't intersect any other subset in more than 2 elements. Since existing subsets include all subsets of size 3, which could potentially intersect with a subset of size 4 in 3 elements. Therefore, it's impossible to include any subset of size 4 or higher if we already have all subsets of size 3. Therefore, the family of all subsets of size ≤3 satisfies the condition, with 175 subsets.But the problem statement says "B₁, B₂, ..., B_j are a family of non-empty subsets of A, when i ≠ j, |B_i ∩ B_j| ≤ 2, find the maximum value of k."Wait, does this mean that the maximum k is 175? But that seems too straightforward. Maybe I'm misunderstanding the problem.Wait, the original problem is in Chinese, translated as:20.1.1 ⋆⋆ Let A = {0,1,2,…,9}. Let B₁, B₂, …, B_k be a family of non-empty subsets of A. When i ≠ j, |B_i ∩ B_j| ≤ 2. Find the maximum value of k.So, according to my previous reasoning, if we take all non-empty subsets of size at most 3, then any two subsets have intersection size at most 2, and the number of such subsets is C(10,1) + C(10,2) + C(10,3) = 10 + 45 + 120 = 175. Therefore, the maximum k is 175. However, maybe there's a mistake here.Wait, but wait, in this case, if we take all subsets of size 1, 2, and 3, then two different subsets of size 3 can intersect in 2 elements, which is allowed. A subset of size 3 and a subset of size 2 can intersect in 2 elements (if the size 2 subset is a subset of the size 3 subset). Similarly, a subset of size 3 and a subset of size 1 can intersect in 1 element. Therefore, all intersections are within the allowed limit. So this family indeed satisfies the condition.But is there a larger family? For example, can we include some subsets of size 4 without violating the condition? Let's see. Suppose we want to add a subset of size 4, say {0,1,2,3}. Now, we need to check that this subset doesn't intersect any existing subset in more than 2 elements. The existing subsets include all subsets of size 1, 2, 3. So, the subset {0,1,2,3} intersects with the subset {0,1,2} (size 3) in 3 elements, which is not allowed. Therefore, we cannot add any subset of size 4 if we have already included all subsets of size 3. Hence, if we want to include subsets of size 4, we have to exclude all subsets of size 3 that intersect them in 3 elements. But since subsets of size 3 are numerous, it's impractical to include subsets of size 4 without violating the intersection condition.Therefore, including all subsets of size ≤3 gives a valid family of 175 subsets. Is there a way to include more subsets beyond 175? For example, maybe some subsets of size 4 that don't conflict with existing subsets. But as shown, it's impossible. Alternatively, maybe removing some subsets of size 3 and adding some subsets of size 4. But I need to check if this could lead to a larger family.Suppose we remove all subsets of size 3 that contain a particular element, say 0, and then add some subsets of size 4 that include element 0. But the problem is that the subsets of size 4 would still intersect with other subsets of size 3 not containing 0 in up to 3 elements. For example, if we add {0,1,2,3}, even if we remove all size 3 subsets containing 0, there are still size 3 subsets like {1,2,4} that intersect with {0,1,2,3} in {1,2}, which is allowed. Wait, actually, maybe if we carefully remove certain subsets of size 3 and add subsets of size 4, we can have a larger family. But this requires careful analysis.Let me attempt a small example. Suppose we remove one subset of size 3, say {0,1,2}, and add a subset of size 4, say {0,1,2,3}. The added subset {0,1,2,3} will intersect with other subsets of size 3 in at most 2 elements. For example, with {0,1,3}, the intersection is {0,1,3} ∩ {0,1,2,3} = {0,1,3}, size 3. Oh, that's a problem. So even if we remove {0,1,2}, adding {0,1,2,3} would conflict with {0,1,3}, which is still in the family. Therefore, it's not possible to just remove one subset and add another.Alternatively, remove all subsets of size 3 that contain {0,1,2}, but that would be multiple subsets. This approach seems messy and might not lead to a net gain in the number of subsets.Alternatively, abandon subsets of size 3 and go for larger subsets with controlled intersections. But this seems difficult.Alternatively, use the concept of a (10, k, 2) code, where each codeword is a subset, and the distance is the symmetric difference. But not sure.Wait, another thought: The family of all subsets of size at most 2, along with all subsets of size 3 that do not contain any pair more than once. Wait, but how?Alternatively, if we can find a family of 3-element subsets where each pair is contained in at most one subset, similar to a Steiner triple system. But a Steiner triple system S(2,3,10) would require that every pair is in exactly one triple. However, such a system doesn't exist because 10 choose 2 is 45, and each triple contains 3 pairs, so the number of triples would be 45/3=15. But 15*3=45, and the total number of triples in a Steiner system would be 15. However, in our case, we can have multiple triples containing the same pair, as long as no two triples share more than two elements. Wait, actually, in a Steiner triple system, two triples intersect in at most one element. But in our problem, two triples can intersect in two elements. So, actually, a Steiner triple system is more restrictive. Therefore, our problem allows more subsets.Therefore, the family of all 3-element subsets is allowed, and as we saw earlier, this gives 120 subsets. But combined with all subsets of size 1 and 2, it's 175. Since the problem allows any subsets as long as their pairwise intersections are ≤2, and since subsets of size ≤3 automatically satisfy that when combined with others, then 175 is the maximum. But I need to confirm.Wait, let's see. If we take a subset S of size 4, then it can't be included with any of its 3-element subsets, because S and a 3-element subset of it would intersect in 3 elements. Therefore, if we want to include any subsets of size ≥4, we have to exclude all their 3-element subsets. But since there are many 3-element subsets, removing them to include a few 4-element subsets would probably result in a net loss. Therefore, it's better to keep all subsets of size ≤3.Therefore, the maximum k is 175. But wait, let me verify with the example of two subsets of size 3: {0,1,2} and {0,1,3}. Their intersection is {0,1}, size 2, which is allowed. Similarly, {0,1,2} and {0,3,4} intersect in {0}, size 1. So, indeed, any two subsets of size 3 intersect in at most 2 elements, so they are okay. And subsets of size 3 with subsets of size 2 can intersect in at most 2 elements. Therefore, the family of all subsets of size ≤3 works, giving k=175.But wait, the problem says "non-empty subsets", so including all subsets from size 1 up to size 3. Then, the total number is C(10,1)+C(10,2)+C(10,3)=10+45+120=175. However, I have a feeling that this might not be the correct answer, because usually, in such problems, the answer is not as straightforward as summing combinations. Maybe there is a restriction I'm missing.Wait, for example, the original problem could be interpreted as a family of subsets where every pair of subsets intersects in at most two elements, but when you include all subsets of size 3, the pairwise intersections are indeed at most two elements. So according to that, 175 is correct. However, in combinatorial literature, the maximum family of subsets of {0,1,...,9} with pairwise intersections at most 2 is known?Wait, I recall that for the set of size n, the maximum family of subsets with pairwise intersections at most t is the sum of combinations C(n,0) + C(n,1) + ... + C(n,t+1). Wait, no, that's the theorem by Sauer, Shelah, and others, but that's for families shattering VC-dimension. Not exactly the same.Wait, another theorem: In the problem of families of sets with pairwise intersections at most t, the maximum size is indeed the sum of C(n,0) + C(n,1) + ... + C(n, t+1). Is that the case?Wait, according to the theorem by Kleitman, the maximum size of a family of subsets of an n-element set with the property that the intersection of any two distinct subsets is at most t is indeed the sum_{i=0}^{t+1} C(n, i). This is known as the Kleitman's theorem.Yes, Kleitman's theorem states that the maximum family of subsets where every pairwise intersection has size at most t is the sum of the first t+1 binomial coefficients. So for our case, t=2, so the maximum family size is C(10,0)+C(10,1)+C(10,2)+C(10,3)=1+10+45+120=176. But wait, the theorem says the sum up to t+1, which would be 0 to 3. But the problem states "non-empty subsets", so we subtract 1 for the empty set. Hence, 176-1=175. Therefore, the answer is 175.Therefore, according to Kleitman's theorem, the maximum k is 175. Therefore, the answer is boxed{175}.But wait, let me confirm Kleitman's theorem. According to Kleitman's theorem, for a family of subsets where every pairwise intersection has size at most t, the maximum size is indeed the sum of the binomial coefficients up to t+1. This is true when the family is closed under taking subsets, which ours is not, but in the case where the family includes all subsets of size up to t+1, then this is the maximum. However, in our case, the problem allows any non-empty subsets, not necessarily closed under inclusion, but the maximum is achieved by taking all subsets of size up to t+1=3. Therefore, the answer is 175.Yes, that seems to be correct.Final Answerboxed{175}</think>Given the set ( A = {0, 1, 2, ldots, 9} ), we need to find the maximum number ( k ) of non-empty subsets ( B_1, B_2, ldots, B_k ) such that the intersection of any two distinct subsets ( B_i ) and ( B_j ) has at most 2 elements.To solve this, we consider the family of all subsets of ( A ) with sizes 1, 2, and 3. This is because any two subsets of size 3 can intersect in at most 2 elements, and subsets of smaller sizes (1 or 2) will intersect with each other or with subsets of size 3 in at most 2 elements. The number of such subsets is calculated as follows:- The number of subsets of size 1: ( binom{10}{1} = 10 )- The number of subsets of size 2: ( binom{10}{2} = 45 )- The number of subsets of size 3: ( binom{10}{3} = 120 )Summing these, we get the total number of non-empty subsets:[10 + 45 + 120 = 175]According to Kleitman's theorem, the maximum family of subsets where every pairwise intersection has size at most ( t ) is the sum of the binomial coefficients up to ( t+1 ). For ( t = 2 ), this sum includes subsets of sizes 0, 1, 2, and 3. Excluding the empty set, we get the total number of non-empty subsets as 175.Thus, the maximum value of ( k ) is (boxed{175}).,<think>Alright, so I have this problem here: We have the set A = {0,1,2,...,9}, and there's a family of non-empty subsets B₁, B₂, ..., B_k. The condition is that for any two different subsets B_i and B_j, the size of their intersection is at most 2. The question is asking for the maximum possible value of k, which means I need to find how many such subsets I can have without any two of them sharing more than two elements.Hmm, okay. Let me start by understanding the problem better. We have a universe of 10 elements (0 through 9), and we need to create as many subsets as possible where each subset is non-empty, and any two subsets intersect in at most 2 elements. So, no two subsets can share three or more elements. The goal is to maximize the number of such subsets.This reminds me of combinatorial design problems, where you have certain intersection properties between subsets. Maybe something like a block design? In block design theory, you have parameters like v (number of elements), k (size of each block), λ (number of blocks that contain any given pair), etc. But here, the problem is a bit different because we're not fixing the size of the subsets; they can be any non-empty size, as long as the intersection condition is satisfied.Alternatively, this is similar to the concept of a family of sets with bounded pairwise intersections. There's a theorem called the Fisher's inequality or maybe the Erdős–Ko–Rado theorem, which gives maximum family sizes under intersection conditions. Let me recall.The Erdős–Ko–Rado theorem primarily deals with families of subsets where each subset has a fixed size, and intersections are at least a certain size. But in our case, the subsets can be of any size, and the intersections are bounded above. So maybe it's a different theorem or a variation.Another thought: if we can model this as a graph problem, where each subset is a vertex, and edges connect subsets whose intersection is more than 2. Then the problem reduces to finding the maximum independent set in this graph. However, maximum independent set is generally hard to compute, but maybe with the structure here, we can find some bounds.But perhaps a better approach is to use combinatorial bounds like the Fisher's inequality or use double counting or inclusion-exclusion principles. Let's think step by step.First, note that each subset B_i is a non-empty subset of A, which has 10 elements. So the total number of possible subsets is 2¹⁰ - 1 = 1023. But of course, we can't have all of them because many pairs would intersect in more than two elements.We need to pick as many subsets as possible where each pair intersects in at most two elements. To maximize k, we need to arrange these subsets so that they don't "overlap" too much.One strategy might be to use subsets of size 3. If all subsets are size 3, then any two subsets can intersect in at most 2 elements, which satisfies the condition. But if we use larger subsets, say size 4, then two subsets could intersect in 3 elements, which would violate the condition. So maybe restricting subset sizes to 3 would help. However, subsets of size 1 or 2 can also be included, as their intersections with other subsets would be smaller. For example, a singleton set can only intersect any other set in at most 1 element, so it's safe. Similarly, a pair (subset of size 2) can intersect another subset in at most 2 elements. Wait, but if we have another subset that contains that pair, their intersection would be 2 elements, which is allowed. So perhaps a mixture of subsets of different sizes, as long as they don't violate the intersection condition.But perhaps the maximum number is achieved when all subsets are of size 3. Let me check. How many 3-element subsets are there in a 10-element set? That's C(10,3) = 120. If we use all 3-element subsets, then any two subsets intersect in at most 2 elements, which satisfies the condition. So, 120 subsets. But can we do better by including some subsets of other sizes?For example, suppose we include all subsets of size 1, 2, and 3. The total number of these is C(10,1) + C(10,2) + C(10,3) = 10 + 45 + 120 = 175. However, the problem is that some of these subsets might intersect in more than two elements. Wait, no. Subsets of size 1 can only intersect other subsets in at most 1 element. Subsets of size 2 can intersect other subsets of size 2 in at most 2 elements (if they are the same), but since all subsets are distinct, two different subsets of size 2 intersect in at most 1 element. Similarly, a subset of size 2 and a subset of size 3 can intersect in at most 2 elements. A subset of size 1 and any other subset can intersect in at most 1 element.Wait a second, actually, if we include subsets of size 1, 2, and 3, then any two different subsets will have intersections of size at most 2. Because:- Two different singletons: intersect in 0 elements.- A singleton and a pair: intersect in 0 or 1 element.- A singleton and a 3-element subset: intersect in 0 or 1 element.- Two different pairs: intersect in 0 or 1 element.- A pair and a 3-element subset: intersect in 0, 1, or 2 elements.- Two different 3-element subsets: intersect in 0, 1, or 2 elements.So, in all cases, the intersections are at most 2. Therefore, including all subsets of size 1, 2, and 3 would give us 175 subsets, and all of them satisfy the intersection condition. Wait, but that contradicts my initial thought that the problem was difficult. Maybe I'm missing something here.Wait, the problem states "non-empty subsets", so indeed, if all subsets of size 1, 2, 3 are allowed, and they all satisfy the pairwise intersection condition, then 175 is the answer. But that seems too straightforward. Let me check the problem again.The problem says: "B₁, B₂, ..., B_j are a family of non-empty subsets of A, when i ≠ j, |B_i ∩ B_j| ≤ 2, find the maximum value of k."Hmm, so if the family includes all subsets of size 1, 2, 3, then indeed, the pairwise intersections are at most 2, so the maximum k would be 10 + 45 + 120 = 175. But maybe this is correct? Let me verify with an example.Take two subsets of size 3: {0,1,2} and {0,1,3}. Their intersection is {0,1}, which is size 2. Okay. If I take a subset of size 2, say {0,1}, and a subset of size 3, say {0,1,2}, their intersection is {0,1}, which is size 2. A subset of size 1, say {0}, and a subset of size 2, {0,1}, their intersection is {0}, size 1. So yes, all intersections are indeed ≤2. So, why is the problem marked with two stars (indicating it's difficult)? Maybe there's a mistake in my reasoning.Wait, wait. Wait a second. If I include subsets of size larger than 3, say size 4, then two subsets of size 4 could intersect in 3 elements, which would violate the condition. So to be safe, if we restrict ourselves to subsets of size at most 3, then all intersections are okay. But if we include subsets larger than size 3, then we have to be careful. However, in the family of all subsets of size 1, 2, 3, there are no subsets larger than 3, so all intersections are indeed okay. So, according to this, the maximum k is 175.But that seems too high. Let me check the total number of subsets. The total number of non-empty subsets of A is 2¹⁰ -1 = 1023. If we take all subsets of size ≤3, that's 175. But the problem is asking for the maximum k where any two subsets intersect in at most two elements. But if the answer is 175, that's a lot, but maybe it's correct?Wait, but perhaps there is a mistake here. Wait, if two subsets are both of size 3, they can intersect in up to 2 elements. If they are both of size 2, they can intersect in up to 1 element. If one is size 3 and the other is size 2, they can intersect in up to 2 elements. So, all pairs are okay. So, why can't we take all subsets of size at most 3? Is there a restriction in the problem that I'm missing?Wait, the problem says "non-empty subsets of A", so they can be any size from 1 to 10. But if we take all subsets of size at most 3, then we satisfy the intersection condition. So, why isn't the answer 175? Maybe I'm missing a key point here.Wait, let's check if two subsets of size 3 can actually intersect in 3 elements. For example, if we have subsets {0,1,2} and {0,1,2}, but since they are the same subset, but in the family, we are considering distinct subsets. So, as long as all subsets are distinct, two different subsets of size 3 can intersect in at most 2 elements. Similarly for other sizes.Therefore, the family of all subsets of size 1, 2, and 3 should satisfy the condition, and the number is 175. However, maybe this is not the maximum? Maybe we can include some subsets of larger size as well, as long as they don't intersect too much with others. For example, if we include a subset of size 4, but ensure that it doesn't share more than two elements with any other subset. But how?Suppose we include a subset S of size 4. Then, any other subset in the family must not share three elements with S. So, we would have to remove all subsets that contain three elements from S. But this might complicate things. Alternatively, if we can find a way to include some larger subsets without violating the intersection condition, then maybe we can have a larger family. But it's not obvious how to do that.Alternatively, maybe the answer is indeed 175. However, in combinatorial literature, there is a concept called a "clutter" where no subset contains another, but this is different. Alternatively, in coding theory, binary codes with certain distance properties can be analogous to set systems with intersection constraints.Wait, another approach: Let's think of each subset as a vector in a 10-dimensional vector space over the binary field, where each coordinate corresponds to an element of A, and a 1 indicates presence in the subset. Then, the intersection size of two subsets is the dot product of their vectors. So, the condition is that the dot product (over integers) is at most 2. However, I don't know if this helps directly, but maybe bounds from coding theory can apply here.In coding theory, the maximum number of binary codes with length n and certain distance properties is studied. However, the dot product here isn't exactly the Hamming distance. But maybe the concept of a constant-weight code could be related. Wait, if we require that the intersection (which is like the inner product) is bounded, maybe we can use the Johnson bound or something similar. But I'm not sure.Alternatively, maybe use the inclusion-exclusion principle to bound the maximum number of subsets. Let me consider the total number of pairs of elements in A. There are C(10,2) = 45 pairs. Each subset of size m contains C(m,2) pairs. If we have k subsets, then the total number of pairs covered by all subsets is the sum over each subset of C(|B_i|, 2). However, since any two subsets share at most 2 elements, which means that any pair of elements can be contained in at most... hmm, how many subsets? Wait, if two subsets share a pair of elements, then their intersection is at least 2, but the problem states that the intersection is at most 2. Wait, no. If two subsets share a pair, their intersection is 2. If they share more, then it's more. But the problem says that the intersection is at most 2 for any two subsets.Therefore, any pair of elements can be contained in at most one subset. Wait, is that true? Let me see.Suppose a pair {a,b} is contained in two different subsets, say B₁ and B₂. Then, |B₁ ∩ B₂| ≥ 2, which is allowed as long as it's exactly 2. But if B₁ and B₂ also share another element, say c, then their intersection is at least {a,b,c}, which is 3, violating the condition. Therefore, to ensure that any two subsets share at most 2 elements, we must ensure that any pair {a,b} is contained in at most one subset. Wait, no, actually, if two subsets share a pair {a,b}, but no other elements, then their intersection is 2, which is okay. But if another subset also contains {a,b}, then each pair of subsets containing {a,b} would intersect in 2 elements. Wait, but if B₁ contains {a,b,c}, B₂ contains {a,b,d}, and B₃ contains {a,b,e}, then each pair of these subsets intersects in {a,b}, which is size 2. So, in this case, the pair {a,b} is contained in three subsets, but the intersections between those subsets are still size 2. Therefore, actually, a pair can be in multiple subsets, as long as the subsets don't share more than that pair. So, the earlier statement that a pair can be in at most one subset is not necessary. Therefore, my previous reasoning was incorrect.Wait, so here's the thing: If two subsets share a pair {a,b}, then their intersection is exactly 2, which is allowed. So, multiple subsets can contain the same pair {a,b}, provided that no two subsets share three or more elements. So, the same pair can be in multiple subsets. Therefore, the constraint is not on how many times a pair appears, but rather on how much any two subsets overlap. So, the pair can be in multiple subsets, as long as those subsets don't share another element.But if we have a pair {a,b} in subsets B₁, B₂, B₃, etc., then each of these subsets can have other elements, but they can't share any other element with each other. For example, if B₁ = {a,b,c}, B₂ = {a,b,d}, B₃ = {a,b,e}, then each pair of B₁, B₂, B₃ intersect in exactly {a,b}, which is allowed. However, if another subset B₄ = {a,b,c,d}, then B₄ intersects B₁ in {a,b,c}, which is size 3, which is not allowed. So, if we have a pair {a,b} in multiple subsets, those subsets must not share any other element with each other.Therefore, in this case, the number of subsets containing the pair {a,b} is limited by the number of remaining elements. Since A has 10 elements, if we have a pair {a,b}, there are 8 remaining elements. So, we can have at most 8 subsets containing {a,b}, each adding a distinct remaining element. However, if we want subsets larger than size 3, say size 4, containing {a,b}, then each such subset would need to include two more elements. But then, two such subsets could share one element, leading to an intersection of {a,b,c}, which is size 3. So, actually, if we have a pair {a,b}, the number of subsets containing {a,b} is limited by how many other elements we can add without overlapping.Wait, this is getting complicated. Maybe it's better to stick with the initial idea of using all subsets of size at most 3. Since in that case, even if multiple subsets contain the same pair, their intersections are still size 2. For example, {a,b}, {a,b,c}, {a,b,d} all pairwise intersect in {a,b} or {a}, which is within the limit. Wait, {a,b} and {a,b,c} intersect in {a,b}, which is size 2. {a,b,c} and {a,b,d} intersect in {a,b}, which is size 2. {a,b} and {a,b,d} intersect in {a,b}, size 2. So, actually, even subsets of different sizes containing the same pair are okay.Therefore, maybe including all subsets of size 1, 2, 3 is allowed and gives 175 subsets. But perhaps there is a way to include more subsets by cleverly adding some larger subsets without violating the intersection condition. However, since larger subsets risk overlapping too much with others, it's tricky. Let's consider.Suppose we try to add a subset of size 4, say {a,b,c,d}. Now, we have to ensure that this subset doesn't intersect any existing subset in more than 2 elements. The existing subsets are all subsets of size 1, 2, 3. So, the intersections with subsets of size 1: at most 1 element, which is okay. The intersections with subsets of size 2: the subset {a,b,c,d} can intersect a size 2 subset in at most 2 elements (if the size 2 subset is a subset of {a,b,c,d}). Similarly, intersections with size 3 subsets: the subset {a,b,c,d} can intersect a size 3 subset in up to 3 elements, which would be a problem. For example, if there is a subset {a,b,c}, then intersecting with {a,b,c,d} gives {a,b,c}, which is size 3, violating the condition. Therefore, we cannot include any subset of size 4 if there are existing subsets of size 3 that are subsets of it.Therefore, to add a subset of size 4, we need to remove all subsets of size 3 that are contained within it. For example, if we add {a,b,c,d}, we need to remove the four subsets {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}. Each of these subsets is of size 3 and would intersect the size 4 subset in 3 elements. So, by adding one subset of size 4, we have to remove 4 subsets of size 3. Is this a good trade-off? Let's see: we lose 4 subsets but gain 1. So, the total number decreases by 3. Therefore, this is not beneficial.Similarly, if we try to add a subset of size 5, we would have to remove all subsets of size 3, 4 (if any), that are contained within it. But since we don't have subsets of size 4, we would have to remove C(5,3) = 10 subsets of size 3. So, removing 10 subsets to add 1, which is a net loss of 9. Not helpful.Alternatively, if we remove some subsets of size 3 and replace them with larger subsets. For example, if we have a collection of subsets where each larger subset "covers" multiple smaller subsets, but without overlapping too much with others. But this seems complex and might not result in a net gain.Alternatively, maybe we can find a different family of subsets that includes some larger subsets but still maintains the intersection condition and has a larger total number than 175. However, I can't see an obvious way to do that. It might require more advanced combinatorial techniques.Wait, another angle: The problem is similar to what is called a "2-intersecting family", but actually, it's the opposite. Here, we want all pairs to intersect in at most 2 elements, which is a upper bound on intersection size. In literature, such families are sometimes called "pairwise intersecting families with bounded intersection". There might be known results about the maximum size of such families.Upon a quick recall, for a family of subsets where the intersection of any two distinct subsets is at most t, the maximum size is equal to the sum of binomial coefficients from 0 to t+1. Wait, no, that's not exactly. Wait, actually, if we consider all subsets of size ≤ t+1, then any two subsets have intersection at most t. Wait, let me check.Suppose we have subsets of size at most t+1. Then, the intersection of any two subsets is at most t+1, but actually, if two subsets have size t+1, their intersection can be up to t+1. So, to have intersections at most t, we need subsets of size at most t. Because if subsets have size t, their intersection can be at most t. Wait, no. For example, two subsets of size t can have intersection up to t. So, if we require that any two subsets have intersection at most t, then the maximum family is all subsets of size at most t. But in our problem, t = 2, so all subsets of size at most 2. Wait, but that would be C(10,1) + C(10,2) = 10 + 45 = 55 subsets. But in our problem, the answer was 175 when including subsets of size 3. So, this contradicts.Wait, perhaps the reasoning is different. If we require that any two subsets have intersection at most t, then all subsets of size greater than t can be included only if they don't share t+1 elements with any other subset. But this seems complicated.Wait, actually, in our problem, the family can include subsets of any size, as long as no two subsets share more than 2 elements. So, subsets of size 3 can be included because two different size 3 subsets can share at most 2 elements. Similarly, subsets of size 4 can be included if they don't share 3 elements with any other subset. But as we saw earlier, including a subset of size 4 requires removing all subsets of size 3 that are subsets of it, which might not be beneficial.So, if we stick to the family of all subsets of size at most 3, we get 175 subsets. But if we try to include some subsets of size 4, 5, etc., we might have to remove more subsets than we add, resulting in a smaller total. Therefore, perhaps 175 is indeed the maximum.But wait, in the example earlier, if we include all subsets of size 1, 2, 3, then two different subsets of size 3 can intersect in 2 elements, which is allowed, so we don't have to remove any subsets. So why is this possible? Because even though two subsets of size 3 can intersect in 2 elements, that's acceptable. So the total number of subsets is 175, and this satisfies the condition.But if that's the case, then why is this problem considered to be of medium difficulty (two stars)? Maybe the answer is not 175. Wait, but let's check some references.Wait, another thought: Maybe the problem is a variation of the problem where you have to find the maximum family of subsets where every two subsets intersect in at most λ elements. For λ=2, in a 10-element set, what is the maximum family?According to the inclusion properties, if you allow all subsets of size ≤3, then you get a family where every two subsets intersect in ≤2 elements. However, is this the maximum possible?Wait, let's suppose that there exists a larger family. For example, if we can include some subsets of size 4 without removing too many subsets of size 3. But as we saw earlier, adding one subset of size 4 would require removing 4 subsets of size 3. So, unless there is a way to add multiple subsets of size 4 without overlapping in their 3-element subsets, which seems difficult.Alternatively, maybe use projective planes or other combinatorial structures to arrange subsets such that their intersections are controlled. But projective planes usually require specific parameters, and I'm not sure if they apply here.Alternatively, use the probabilistic method to estimate the maximum family size. But this might be complicated.Alternatively, think about the problem in terms of graph theory. Each subset is a vertex, and edges connect subsets that intersect in more than 2 elements. Then, the problem reduces to finding the maximum clique in the complement graph. However, finding maximum cliques is NP-hard, and I don't know the structure of this graph.Alternatively, use the theorem of de Bruijn and Tengbergen, which states that for families of subsets of [n] with pairwise intersections of size at most t, the maximum family is the union of all subsets of size at most t+1. Wait, but in our case, t=2. So, according to this theorem, the maximum family would be all subsets of size at most 3, which is 175. So, is this the case?Wait, actually, I need to check the exact statement. The de Bruijn–Tengbergen theorem actually concerns families where every two subsets have intersections exactly t, but I might be misremembering. Let me check.Wait, no. The de Bruijn–Tengbergen theorem is about the maximum size of a family of subsets with the property that any two subsets have the same intersection size. But in our problem, the intersection can be at most 2, not exactly 2. So, it's a different problem.Wait, maybe the answer is indeed 175, but perhaps the problem is in Chinese, and the original problem might have a different condition, but according to the translation provided, it's intersections at most 2, so the answer is 175. However, in the original problem statement, if the user mentioned "B₁, B₂, ..., B_j are a family of non-empty subsets of A, when i ≠ j, |B_i ∩ B_j| ≤ 2, find the maximum value of k", then according to standard combinatorial results, the maximum k is indeed the number of subsets of size at most 3, which is 175.But wait, in the example I considered earlier, if we have all subsets of size at most 3, then two subsets can intersect in 0, 1, or 2 elements, all of which are allowed. So, why isn't this the maximum? Because you can't have more subsets without forcing some intersection to exceed 2. Suppose we try to add a subset of size 4. As mentioned, this would require that none of its 3-element subsets are present in the family. But in our current family, all 3-element subsets are present, so we can't add any subset of size 4. Therefore, 175 is indeed the maximum.But I need to verify this with another approach. Let's use double counting.Suppose we have a family F of subsets of A, each non-empty, with pairwise intersections at most 2. Let's denote the number of subsets in F as k. We need to find the maximum k.Consider the number of ordered triples (B, C, x) where B and C are distinct subsets in F, and x is an element of their intersection. Each such triple is counted once. The total number of such triples is equal to the sum over all pairs B, C of |B ∩ C|. Since for each pair B ≠ C, |B ∩ C| ≤ 2, the total number of triples is ≤ 2 * C(k, 2).On the other hand, let's count the number of such triples in another way. For each element x in A, let's denote by d_x the number of subsets in F that contain x. Then, the number of triples (B, C, x) where x ∈ B ∩ C is equal to the sum over x of C(d_x, 2). Therefore, we have:Sum_{x ∈ A} C(d_x, 2) ≤ 2 * C(k, 2)This inequality comes from double-counting the triples.Let me write that down:Σ_{x=0}^9 [C(d_x, 2)] ≤ 2 * C(k, 2)Which simplifies to:Σ_{x=0}^9 [d_x(d_x - 1)/2] ≤ 2 * [k(k - 1)/2]Multiplying both sides by 2:Σ_{x=0}^9 [d_x(d_x - 1)] ≤ 2 * k(k - 1)This is an important inequality. Let's denote this as Inequality (1).Additionally, we know from the principle of inclusion that:Σ_{x=0}^9 d_x = Σ_{B ∈ F} |B|.Let's denote S = Σ_{B ∈ F} |B|. Then, Σ_{x=0}^9 d_x = S.Our goal is to relate S and k to get a bound on k.Applying Cauchy-Schwarz inequality to the sum Σ_{x=0}^9 d_x^2.We know that Σ_{x=0}^9 d_x^2 ≥ (Σ_{x=0}^9 d_x)^2 / 10 = S² / 10.From Inequality (1):Σ d_x² - Σ d_x ≤ 2k(k - 1)But Σ d_x = S, so:Σ d_x² ≤ 2k(k - 1) + SBut Σ d_x² ≥ S² / 10, so:S² / 10 ≤ 2k(k - 1) + SRearranging:S² - 10S - 20k(k - 1) ≤ 0This is a quadratic in S:S² - 10S - 20k(k - 1) ≤ 0Solving for S, we can find that the discriminant must be non-negative:100 + 80k(k - 1) ≥ 0Which is always true. But perhaps we can bound S in terms of k.Alternatively, let's consider the case when all subsets are of size at most 3. Then, S = Σ |B|. For the family F containing all subsets of size 1, 2, 3:S = C(10,1)*1 + C(10,2)*2 + C(10,3)*3 = 10*1 + 45*2 + 120*3 = 10 + 90 + 360 = 460.Then, Σ d_x = 460. Each d_x is the number of subsets containing element x. Since each element is in C(9,0) + C(9,1) + C(9,2) subsets of size 1, 2, 3. Wait, no. For each element x, the number of subsets containing x is:- Size 1: 1 (the subset {x})- Size 2: C(9,1) = 9 (pair {x, y} for each y ≠ x)- Size 3: C(9,2) = 36 (triple {x, y, z} for each pair y, z ≠ x)Therefore, total d_x = 1 + 9 + 36 = 46 for each x. Since there are 10 elements, Σ d_x = 10*46 = 460, which matches the earlier calculation.Then, Σ d_x² = 10*(46)^2 = 10*2116 = 21160.Now, according to Inequality (1):Σ d_x(d_x - 1) = Σ d_x² - Σ d_x = 21160 - 460 = 20700 ≤ 2k(k - 1)But in this case, k = 175, so 2k(k - 1) = 2*175*174 = 2*30450 = 60900. So, 20700 ≤ 60900, which holds. Therefore, this family satisfies the inequality, but the inequality is not tight here.This suggests that there could be a larger family that still satisfies the inequality, but how?Wait, perhaps if we can arrange subsets such that the degrees d_x are more evenly distributed, allowing for a higher k. However, in the case of all subsets of size ≤3, the degrees d_x are all equal (each element is in 46 subsets), which is already quite efficient.Alternatively, suppose we have a different family where degrees d_x are higher, but Σ d_x² is lower. However, by Cauchy-Schwarz, the minimal Σ d_x² occurs when all d_x are equal. Therefore, the family with all subsets of size ≤3 already minimizes Σ d_x² for a given S, meaning that this family is optimal in terms of the left-hand side of Inequality (1). Therefore, if we want to maximize k, we need to have a larger S, but subject to the constraint from Inequality (1). But since in this case, the left-hand side (20700) is much smaller than the right-hand side (60900), there's a lot of slack.This suggests that maybe a larger family is possible. However, in reality, we cannot have a larger family because when you include more subsets, especially larger ones, you might have to start increasing the pairwise intersections beyond 2. But according to the double counting argument, the inequality is not tight, so perhaps the bound is not useful here.Alternatively, perhaps use the Fisher-type inequality. Wait, in a block design where each pair of blocks intersects in at most λ elements, then the maximum number of blocks is related to the number of elements. But I don't recall the exact relation.Alternatively, think about the problem in terms of the inclusion of subsets. For each element, the number of subsets containing it is d_x. If all d_x are equal, then as in our previous case, d_x = 46. If we try to make a larger family, we have to include more subsets, which would require increasing some d_x. However, increasing d_x would increase Σ d_x², which affects the inequality. But since the inequality is not tight, perhaps there is room.Wait, but let's test with a larger k. Suppose k = 175 + m, where m is the number of additional subsets we can include. Let's say we add m subsets of size 4, each requiring the removal of 4 subsets of size 3. So, net gain is m - 4m = -3m, which is negative. So, no gain.Alternatively, perhaps if we can add some subsets of size 4 that don't require removing existing subsets. But as mentioned earlier, a subset of size 4 would intersect any of its supersets of size 3 in 3 elements, which are already present in the family. Therefore, we cannot have such subsets.Alternatively, if we don't include all subsets of size 3, but only some of them, then we might have space to include larger subsets. For example, if we have an incomplete family of subsets of size 3, such that no two subsets share three elements with a larger subset. However, constructing such a family is non-trivial and likely would not surpass the total number 175.Alternatively, think about the problem from the perspective of the Erdos-Ko-Rado theorem, which gives the maximum family of subsets of size k where every two subsets intersect. But in our case, the intersection is bounded above, not below, and subsets can be of any size. So, not directly applicable.Another approach: Use the concept of a hypergraph. We are looking for a 3-uniform hypergraph (if we stick to subsets of size 3) with pairwise edges intersecting in at most 2 vertices. But in hypergraph terminology, this is just a 3-uniform hypergraph, which naturally has edges intersecting in at most 2 vertices. However, the maximum number of edges in a 3-uniform hypergraph on 10 vertices is C(10,3)=120, which is what we have. If we include also 2-edges and 1-edges, the total is 175. But can a hypergraph with edges of size 1, 2, and 3 have more than 175 edges if we allow some edges of larger size? The problem is that larger edges would intersect the 3-edges in more than 2 vertices, so we cannot.Alternatively, if we consider only linear hypergraphs, where any two edges intersect in at most one vertex. But in our problem, the intersection can be up to 2, which is a different condition.Wait, but in our problem, the hypergraph can have edges of any size, with pairwise intersections at most 2. So, edges can be larger, but must not share too many elements. The maximum number of edges in such a hypergraph is what we need.According to a theorem by Kővári–Sós–Turán, which gives upper bounds on the number of edges in hypergraphs with certain intersection properties, but I'm not sure of the exact formulation for our case.Alternatively, use the probabilistic method: Calculate the expected number of subsets we can choose randomly without conflicts, but this might not give a precise answer.Alternatively, refer to known results: A quick search in my mind tells me that for a family of subsets of an n-element set where every two subsets intersect in at most t elements, the maximum family size is indeed the sum of binomial coefficients from 0 to t+1. For n=10 and t=2, this would be C(10,0)+C(10,1)+C(10,2)+C(10,3) = 1 +10+45+120=176. Wait, but we excluded the empty set, so it's 176 -1 = 175, which matches our previous count. However, some references say that this is the maximum, but I need to verify.Wait, but actually, the family including all subsets of size ≤3 and the empty set would have size 176, but since the problem requires non-empty subsets, it's 175. Therefore, if this is a known result, then the maximum k is 175.But in the problem statement, it's specified that the subsets are non-empty, so 175 is the answer. Therefore, after all this reasoning, I think the maximum value of k is 175.Final Answerboxed{175}</think>Given the set ( A = {0, 1, 2, ldots, 9} ), we need to find the maximum number ( k ) of non-empty subsets ( B_1, B_2, ldots, B_k ) such that the intersection of any two distinct subsets ( B_i ) and ( B_j ) has at most 2 elements.1. Understanding the Problem: - Each subset must be non-empty. - Any two distinct subsets can intersect in at most 2 elements.2. Initial Considerations: - Subsets of size 1, 2, or 3 naturally satisfy the intersection condition because their intersections cannot exceed 2 elements. - Larger subsets (size 4 or more) risk intersecting other subsets in more than 2 elements, which would violate the condition.3. Counting Subsets: - The number of subsets of size 1 is ( binom{10}{1} = 10 ). - The number of subsets of size 2 is ( binom{10}{2} = 45 ). - The number of subsets of size 3 is ( binom{10}{3} = 120 ).4. Total Subsets of Size 1, 2, or 3: - Summing these, we get ( 10 + 45 + 120 = 175 ).5. Verification: - Subsets of size 1, 2, or 3 intersect in at most 2 elements. - Including subsets of larger sizes would require removing smaller subsets to avoid intersections exceeding 2 elements, which does not increase the total count.6. Conclusion: - The maximum number of subsets satisfying the condition is achieved by including all subsets of size 1, 2, and 3.Thus, the maximum value of ( k ) is (boxed{175}).]

🔑:Parse the number of non-empty subsets containing at most three elements:mathrm{C}_{10}^{1}+mathrm{C}_{10}^{2}+mathrm{C}_{10}^{3}=175 text { subsets. }Let C be the family of subsets that satisfy the condition. If |B| geq 4, then B and B-{b}(b in B) cannot both be contained in C. By replacing B with B-{b}, the number of elements in C remains unchanged. Similarly, for all sets B in C with at least 4 elements, perform the corresponding replacements. After these replacements, each set in the family C will be a non-empty set with at most 3 elements. Therefore, the maximum value of k is 175.