🧠:[<think>Okay, so I have this problem with two parts, I and II. Let me start with part I. It says when a is -1, solve the inequality f(x) ≥ 4 - |2x - 1|. Alright, let's break this down step by step.First, f(x) is given as |x - a|. Since a is -1 here, that becomes |x - (-1)|, which simplifies to |x + 1|. So f(x) = |x + 1|. The inequality we need to solve is |x + 1| ≥ 4 - |2x - 1|.Hmm, okay. So we have an absolute value inequality on the left and another absolute value expression subtracted from 4 on the right. To solve this, I might need to consider different cases based on the critical points where the expressions inside the absolute values change their sign. Let's find those critical points.For |x + 1|, the critical point is at x = -1. For |2x - 1|, the expression inside becomes zero when 2x - 1 = 0 => x = 1/2. So the critical points are x = -1 and x = 1/2. These divide the real line into three intervals: (-∞, -1), [-1, 1/2), and [1/2, ∞). I need to consider each interval separately and solve the inequality accordingly.Let me start with the first interval: x < -1.In this case, x + 1 is negative, so |x + 1| = -(x + 1) = -x -1. Also, 2x - 1 would be less than 2*(-1) -1 = -3, so negative. Therefore, |2x - 1| = -(2x - 1) = -2x +1. So the inequality becomes:- x -1 ≥ 4 - (-2x +1) => -x -1 ≥ 4 +2x -1 => -x -1 ≥ 3 + 2xNow, let's solve this:- x -1 -3 -2x ≥ 0 => -3x -4 ≥ 0 => -3x ≥ 4 => x ≤ -4/3.But we are in the interval x < -1. So x ≤ -4/3 is within x < -1 because -4/3 ≈ -1.333, which is less than -1. So in this interval, the solution is x ≤ -4/3.Next interval: -1 ≤ x < 1/2.Here, x + 1 is non-negative, so |x + 1| = x +1. 2x -1 is still negative because x < 1/2, so |2x -1| = -2x +1. So the inequality becomes:x +1 ≥ 4 - (-2x +1) => x +1 ≥ 4 +2x -1 => x +1 ≥ 3 +2xSolving:x +1 -3 -2x ≥ 0 => -x -2 ≥ 0 => -x ≥ 2 => x ≤ -2.But wait, in this interval, x is between -1 and 1/2. But x ≤ -2 contradicts this interval. So in this interval, there is no solution.Third interval: x ≥ 1/2.Here, x +1 is positive, so |x +1| = x +1. 2x -1 is non-negative since x ≥1/2, so |2x -1| =2x -1. The inequality becomes:x +1 ≥ 4 - (2x -1) => x +1 ≥4 -2x +1 =>x +1 ≥5 -2xSolving:x +1 -5 +2x ≥0 =>3x -4 ≥0 =>3x ≥4 =>x ≥4/3.But in this interval, x ≥1/2. So x ≥4/3 (≈1.333) is the solution here.Therefore, combining the solutions from all intervals:x ≤ -4/3 or x ≥4/3.Wait, but let me check this. Let me verify with a test point.For x ≤ -4/3: Let's take x = -2.Left side: | -2 +1 | = | -1 | =1Right side:4 - |2*(-2)-1| =4 - | -5 | =4 -5= -1So 1 ≥ -1, which is true. So that works.For x ≥4/3: Take x=2.Left side: |2 +1|=3Right side:4 - |4 -1|=4 -3=13 ≥1, true.Now, check the middle interval where there was no solution. Take x=0.Left: |0 +1|=1Right:4 - |0 -1|=4 -1=31 ≥3? No, that's false. So correct.But wait, what about x= -1?Left: | -1 +1 | =0Right:4 - | -2 -1 | =4 -3=10 ≥1? No. So x=-1 is not a solution.Similarly, x=1/2:Left: |1/2 +1|=3/2Right:4 - |1 -1|=4 -0=43/2 ≥4? No. So x=1/2 isn't a solution.Therefore, the solution set is x ≤ -4/3 or x ≥4/3. So in interval notation, (-∞, -4/3] ∪ [4/3, ∞).Wait, but let me check if x= -4/3 is included.At x= -4/3:Left: | -4/3 +1 | = | -1/3 | =1/3Right:4 - |2*(-4/3) -1| =4 - | -8/3 -3/3 | =4 - | -11/3 | =4 -11/3 =12/3 -11/3=1/3So 1/3 ≥1/3, which is true. So equality holds, so endpoints included.Similarly, x=4/3:Left: |4/3 +1|=7/3Right:4 - |8/3 -1|=4 - |5/3|=4 -5/3=7/37/3 ≥7/3, which is true. So endpoints are included. So the solution is correct.Therefore, part I answer is x ≤ -4/3 or x ≥4/3.Now, part II: If the solution set of f(x) ≤1 is [0,2], prove that g(x) ≥2.Given f(x)=|x -a|, and the solution set of f(x) ≤1 is [0,2]. So we need to find a such that |x -a| ≤1 implies x ∈ [0,2]. Then we need to prove that g(x)=f(x)+f(x+2) is always at least 2.First, let's recall that the solution to |x -a| ≤1 is [a -1, a +1]. Because |x -a| ≤1 implies x is between a -1 and a +1.But the problem says the solution set is [0,2]. Therefore, [a -1, a +1] = [0,2]. Therefore, a -1 =0 and a +1=2. Solving:a -1=0 => a=1Check a +1=2 =>1 +1=2, which is correct. Therefore, a=1.So now, f(x)=|x -1|. Then, g(x)=f(x)+f(x+2)= |x -1| + |x +2 -1|=|x -1| + |x +1|.So g(x)= |x -1| + |x +1|. We need to prove that g(x) ≥2 for all x.Wait, this is a known result. The sum of the distances from x to 1 and from x to -1. But geometrically, for any real number x, the sum of distances to -1 and 1 is at least 2. Because the minimal sum occurs when x is between -1 and 1, and in that case, the sum is exactly 2. For example, if x is between -1 and 1, then |x -1| + |x +1| = (1 -x) + (x +1)=2. If x is outside that interval, say x >1, then |x -1| + |x +1|= (x -1) + (x +1)=2x, which is greater than 2. Similarly, if x < -1, |x -1| + |x +1|= (1 -x) + (-x -1)= -2x, which is also greater than 2 because x is negative. So in all cases, g(x) ≥2. Hence proved.But since the problem asks to prove it, given that the solution set of f(x) ≤1 is [0,2], which gives a=1, so we need to formalize the proof.Alternatively, maybe there's another way. Let's see.Since f(x) = |x -1|, then g(x) = |x -1| + |x +1|. Let's consider different cases.Case 1: x ≥1.Then |x -1|=x -1 and |x +1|=x +1. So g(x)= (x -1) + (x +1)=2x. Since x ≥1, 2x ≥2*1=2. So g(x) ≥2.Case 2: -1 ≤x <1.Here, |x -1|=1 -x and |x +1|=x +1. So g(x)= (1 -x) + (x +1)=2. So g(x)=2.Case3: x < -1.Then |x -1|=1 -x and |x +1|= -x -1. So g(x)= (1 -x) + (-x -1)= -2x. Since x < -1, -2x >2. So g(x) >2.Hence, in all cases, g(x) ≥2. So proved.Therefore, part II is proven.Final Answer(I) The solution is boxed{(-infty, -dfrac{4}{3}] cup [dfrac{4}{3}, infty)}.(II) It is proven that ( g(x) geqslant 2 ) for all ( x ).</think>Given the function ( f(x) = |x - a| ) and ( g(x) = f(x) + f(x + 2) ):(I) When ( a = -1 ), we need to solve the inequality ( f(x) geqslant 4 - |2x - 1| ). 1. Substitute ( a = -1 ) into ( f(x) ), giving ( f(x) = |x + 1| ).2. The inequality becomes ( |x + 1| geq 4 - |2x - 1| ).3. Identify critical points at ( x = -1 ) and ( x = frac{1}{2} ).4. Solve the inequality in three intervals: ( (-infty, -1) ), ( [-1, frac{1}{2}) ), and ( [frac{1}{2}, infty) ).- For ( x < -1 ): - ( |x + 1| = -x - 1 ) and ( |2x - 1| = -2x + 1 ) - Solve ( -x - 1 geq 3 + 2x ) resulting in ( x leq -frac{4}{3} ).- For ( -1 leq x < frac{1}{2} ): - ( |x + 1| = x + 1 ) and ( |2x - 1| = -2x + 1 ) - Solve ( x + 1 geq 3 + 2x ) resulting in no solution.- For ( x geq frac{1}{2} ): - ( |x + 1| = x + 1 ) and ( |2x - 1| = 2x - 1 ) - Solve ( x + 1 geq 5 - 2x ) resulting in ( x geq frac{4}{3} ).Combining the solutions, we get ( x leq -frac{4}{3} ) or ( x geq frac{4}{3} ). The solution set is (boxed{(-infty, -dfrac{4}{3}] cup [dfrac{4}{3}, infty)}).(II) Given the solution set of ( f(x) leq 1 ) is ([0, 2] ):1. The solution set of ( |x - a| leq 1 ) is ([a - 1, a + 1] ). This implies ( a - 1 = 0 ) and ( a + 1 = 2 ), so ( a = 1 ).2. Substitute ( a = 1 ) into ( g(x) ), giving ( g(x) = |x - 1| + |x + 1| ).- For ( x geq 1 ): - ( |x - 1| = x - 1 ) and ( |x + 1| = x + 1 ) - ( g(x) = 2x geq 2 ).- For ( -1 leq x < 1 ): - ( |x - 1| = 1 - x ) and ( |x + 1| = x + 1 ) - ( g(x) = 2 ).- For ( x < -1 ): - ( |x - 1| = 1 - x ) and ( |x + 1| = -x - 1 ) - ( g(x) = -2x > 2 ).Thus, ( g(x) geq 2 ) for all ( x ). Hence, it is proven that ( g(x) geqslant 2 ).,<think>Okay, let's tackle this problem step by step. First, part (I). So we have f(x) = |x - a|, and when a = -1, we need to solve the inequality f(x) ≥ 4 - |2x - 1|. Let me write that down clearly.Given a = -1, so f(x) = |x - (-1)| = |x + 1|. The inequality becomes |x + 1| ≥ 4 - |2x - 1|. Hmm, okay. So I need to solve |x + 1| + |2x - 1| ≥ 4. Wait, is that right? Because if I bring the |2x - 1| to the left side, the inequality becomes |x + 1| + |2x - 1| ≥ 4. That might be easier to handle. Alternatively, I can consider different cases based on the critical points where the expressions inside the absolute values change sign.The critical points for |x + 1| are at x = -1, and for |2x - 1|, it's when 2x - 1 = 0 => x = 0.5. So I should divide the real line into intervals based on these critical points: x < -1, -1 ≤ x < 0.5, and x ≥ 0.5. Then, in each interval, I can remove the absolute value signs appropriately and solve the inequality.Let me start with the first interval: x < -1.In this case, x + 1 < 0, so |x + 1| = -(x + 1) = -x - 1. Similarly, for |2x - 1|, since x < -1, 2x - 1 < 2*(-1) -1 = -3 < 0, so |2x - 1| = -(2x - 1) = -2x + 1.Thus, the inequality becomes (-x -1) + (-2x +1) ≥ 4 => -3x + 0 ≥ 4 => -3x ≥ 4 => x ≤ -4/3.But wait, the interval we are considering is x < -1. So x ≤ -4/3 is part of this interval since -4/3 ≈ -1.333, which is less than -1. Therefore, in this interval, the solution is x ≤ -4/3.Next interval: -1 ≤ x < 0.5.Here, x + 1 ≥ 0, so |x + 1| = x + 1. For |2x - 1|, since x < 0.5, 2x - 1 < 0, so |2x - 1| = -2x + 1. Therefore, the inequality becomes (x + 1) + (-2x +1) ≥ 4 => (-x + 2) ≥ 4 => -x ≥ 2 => x ≤ -2.But in this interval, x is between -1 and 0.5. So x ≤ -2 doesn't overlap with this interval. Therefore, in this interval, there's no solution.Third interval: x ≥ 0.5.Here, x + 1 ≥ 0, so |x + 1| = x +1. Also, 2x - 1 ≥ 0 because x ≥ 0.5, so |2x -1| = 2x -1. The inequality becomes (x +1) + (2x -1) ≥4 => 3x ≥4 => x ≥4/3.Since x ≥0.5, the solutions here are x ≥4/3. 4/3 is approximately 1.333, which is greater than 0.5, so this is valid.So combining the solutions from all intervals: x ≤ -4/3 or x ≥4/3. Therefore, the solution set is (-∞, -4/3] ∪ [4/3, ∞). Let me check if that's correct.Wait, let's test a value in each interval. For example, take x = -2, which is in x < -1. Then f(x) = |-2 +1| = 1, and 4 - |2*(-2) -1| =4 - |-5| =4 -5= -1. So 1 ≥ -1, which is true. But according to our inequality |x+1| + |2x -1| ≥4, let's compute it. | -2 +1 | + |2*(-2) -1| = 1 + 5 =6 ≥4. True. So x=-2 is a solution. Similarly, at x = -4/3, let's compute. | -4/3 +1 | + |2*(-4/3) -1| = | -1/3 | + | -8/3 -1 | = 1/3 + 11/3 =12/3=4. So equality holds here. So x=-4/3 is included. Similarly, x=4/3: |4/3 +1| + |2*(4/3)-1|= |7/3| + |8/3 -1|=7/3 +5/3=12/3=4. Equality holds. For x=2, |2 +1| + |4 -1|=3 +3=6 ≥4. True.So seems correct. Therefore, part (I) solution is x ≤ -4/3 or x ≥4/3. So in boxed form, (-∞, -4/3] ∪ [4/3, ∞). So part (I) done.Now part (II). If the solution set of the inequality f(x) ≤1 is [0,2], prove that g(x) ≥2. Let's parse this.First, f(x) = |x - a|. The inequality |x -a| ≤1 has solution set [a -1, a +1]. But the problem states that the solution set is [0,2]. Therefore, [a -1, a +1] = [0,2]. Therefore, solving for a: a -1=0 and a +1=2. Thus, a=1. Therefore, a=1. So since the solution set is [0,2], we can find that a=1. Therefore, f(x)=|x -1|. Then, g(x)=f(x)+f(x+2)=|x -1| + |x +2 -1|= |x -1| + |x +1|. Wait, no. Wait, f(x+2)=|x +2 -a|. Since a=1, f(x+2)=|x +2 -1|=|x +1|. Therefore, g(x)=|x -1| + |x +1|. So we need to prove that |x -1| + |x +1| ≥2 for all x. Wait, but this is a standard inequality. The sum of distances from x to 1 and from x to -1. The minimal value occurs when x is between -1 and 1, where the sum is equal to 2 (the distance between -1 and 1). For x outside, the sum is greater than 2. Therefore, |x -1| + |x +1| ≥2 for all x. So in this case, since we have a=1, then g(x) is always at least 2, which is what we needed to prove. Let me check again.But let's go step by step. The problem says: If the solution set of f(x) ≤1 is [0,2], prove that g(x) ≥2. So first, we need to find a. Since f(x)=|x -a|. The inequality |x -a| ≤1 implies a -1 ≤x ≤a +1. The solution set is given as [0,2], so we set a -1=0 and a +1=2. Solving both gives a=1. Therefore, a=1. Then, g(x)=f(x) + f(x+2)=|x -1| + |(x+2) -1|=|x -1| + |x +1|. So as mentioned, this is the sum of distances from x to 1 and from x to -1. The minimal value occurs when x is between -1 and 1, where the sum is exactly 2 (since moving from -1 to 1, the total distance covered is 2). For any x outside this interval, the sum is greater than 2. For example, if x >1, then |x -1| + |x +1|= (x -1) + (x +1)=2x ≥2 (since x ≥1). Similarly, if x < -1, then |x -1| + |x +1|= (1 -x) + (-x -1)= -2x ≥2 (since x ≤-1). So indeed, g(x) is always greater than or equal to 2.Therefore, the proof is straightforward once we determine that a must be 1. The key step is recognizing that the solution set [0,2] for |x -a| ≤1 gives a=1. Then, analyzing g(x) as the sum of two absolute values which geometrically represents the sum of distances from x to 1 and -1, which has a minimum value of 2. Hence, g(x) ≥2 for all x.Let me double-check. If a=1, then f(x)=|x -1|. Then, f(x+2)=|x +2 -1|=|x +1|. Therefore, g(x)=|x -1| + |x +1|. Let's pick x=0: | -1 | + |1|=2. x=1: 0 + 2=2. x=-1: 2 +0=2. x=2:1 +3=4. x=-2:3 +1=4. So indeed, the minimum is 2. Therefore, the inequality holds. Thus, part (II) is proven.Final Answer(I) The solution is boxed{(-infty, -dfrac{4}{3}] cup [dfrac{4}{3}, infty)}.(II) It is proven that ( g(x) geq 2 ) for all ( x ).</think>Given the function ( f(x) = |x - a| ) and ( g(x) = f(x) + f(x + 2) ):(I) When ( a = -1 ), we need to solve the inequality ( f(x) geq 4 - |2x - 1| ). Substituting ( a = -1 ), we have ( f(x) = |x + 1| ). The inequality becomes:[|x + 1| geq 4 - |2x - 1|]Rewriting this, we get:[|x + 1| + |2x - 1| geq 4]We consider the critical points ( x = -1 ) and ( x = 0.5 ), dividing the real line into three intervals: ( x < -1 ), ( -1 leq x < 0.5 ), and ( x geq 0.5 ).1. For ( x < -1 ): [ -x - 1 - 2x + 1 geq 4 implies -3x geq 4 implies x leq -frac{4}{3} ]2. For ( -1 leq x < 0.5 ): [ x + 1 - 2x + 1 geq 4 implies -x + 2 geq 4 implies -x geq 2 implies x leq -2 ] No solution in this interval.3. For ( x geq 0.5 ): [ x + 1 + 2x - 1 geq 4 implies 3x geq 4 implies x geq frac{4}{3} ]Combining the solutions, we get ( x leq -frac{4}{3} ) or ( x geq frac{4}{3} ). The solution set is:[boxed{(-infty, -dfrac{4}{3}] cup [dfrac{4}{3}, infty)}](II) Given the solution set of ( f(x) leq 1 ) is ( [0, 2] ), we find ( a ). The inequality ( |x - a| leq 1 ) has solution ( [a - 1, a + 1] ). Thus, ( a - 1 = 0 ) and ( a + 1 = 2 ), giving ( a = 1 ). Now, ( g(x) = |x - 1| + |x + 1| ). This is the sum of distances from ( x ) to 1 and -1. The minimum value occurs when ( x ) is between -1 and 1, giving ( |x - 1| + |x + 1| = 2 ). For other ( x ), the sum is greater than 2. Hence, ( g(x) geq 2 ) for all ( x ).Thus, it is proven that ( g(x) geq 2 ). The final answer is:[boxed{g(x) geq 2}]]

🔑:Solution: (I) When a=-1, the inequality becomes |x+1|+|2x-1|geqslant 4, When xleqslant -1, the original inequality becomes -x-1-2x+1geqslant 4, solving this yields xleqslant - dfrac {4}{3}. When -1 dfrac {1}{2}, the original inequality becomes x+1+2x-1geqslant 4, solving this yields xgeqslant dfrac {4}{3}. Therefore, the solution set of the original inequality is boxed{{x|xleqslant - dfrac {4}{3} text{ or } xgeqslant dfrac {4}{3}}}. (II) To prove: f(x)leqslant 1 means |x-a|leqslant 1, solving this yields a-1leqslant xleqslant a+1, and the solution set of f(x)leqslant 1 is [0,2], Thus, begin{cases} a-1=0 a+1=2end{cases}, solving this yields a=1, hence f(x)=|x-1|. Now, we only need to prove that g(x)=f(x)+f(x+2)geqslant 2. That is, to prove |x-1|+|x+1|geqslant 2. Because |x-1|+|x+1|=|1-x|+|x+1|geqslant |1-x+x+1|=2. Therefore, boxed{g(x)geqslant 2}.

🧠:[<think>Okay, so I need to find the length of the chord that the line x - y + √10 = 0 cuts from the circle defined by the equation x² + y² - 4x - 4y - 1 = 0. Hmm, let's see. I remember there are a few methods to find the chord length when a line intersects a circle. Maybe I can use the formula involving the distance from the center to the line and then apply the Pythagorean theorem? Let me try to recall.First, I should probably rewrite the circle equation in standard form to find its center and radius. The given equation is x² + y² - 4x - 4y - 1 = 0. To convert this into the standard form (x - h)² + (y - k)² = r², I need to complete the square for both x and y terms.Starting with the x terms: x² - 4x. Completing the square, take half of -4, which is -2, square it to get 4. Similarly for the y terms: y² - 4y. Half of -4 is also -2, squared is 4. So, add and subtract these values inside the equation.So, rewriting the equation:x² - 4x + 4 - 4 + y² - 4y + 4 - 4 - 1 = 0That simplifies to:(x - 2)² + (y - 2)² - 8 - 1 = 0Which becomes:(x - 2)² + (y - 2)² = 9So, the circle has center at (2, 2) and radius 3. Got that down.Now, the line in question is x - y + √10 = 0. I need to find the chord length this line cuts from the circle. As I thought earlier, the method involves finding the distance from the center to the line, then using that to compute the chord length.The formula for the distance from a point (h, k) to the line Ax + By + C = 0 is |Ah + Bk + C| / sqrt(A² + B²). Let's apply that here.First, rewrite the line equation in standard form. The line is x - y + √10 = 0. So, A = 1, B = -1, C = √10.The center of the circle is (2, 2). Plugging into the distance formula:Distance = |1*2 + (-1)*2 + √10| / sqrt(1² + (-1)²)Calculate numerator: |2 - 2 + √10| = |0 + √10| = √10Denominator: sqrt(1 + 1) = sqrt(2)So, the distance from center to the line is √10 / sqrt(2) = sqrt(10/2) = sqrt(5) ≈ 2.236. Hmm, but wait, the radius of the circle is 3. Since sqrt(5) is approximately 2.236, which is less than 3, the line does intersect the circle, so there is a chord. Good.Now, to find the chord length, I can use the formula: chord length = 2 * sqrt(r² - d²), where r is the radius and d is the distance from center to the line.Plugging in the numbers: chord length = 2 * sqrt(3² - (sqrt(5))²) = 2 * sqrt(9 - 5) = 2 * sqrt(4) = 2 * 2 = 4.Wait, that seems straightforward. Let me verify if this is correct. Alternatively, maybe I can check by finding the points of intersection between the line and the circle and then calculate the distance between those two points.Let me try that method as a check. To find the points of intersection, solve the system of equations:1) x - y + √10 = 0 => y = x + √102) (x - 2)² + (y - 2)² = 9Substitute y from equation 1 into equation 2:(x - 2)² + ( (x + √10) - 2 )² = 9Simplify:(x - 2)² + (x + √10 - 2)² = 9Let's compute each term.First term: (x - 2)² = x² - 4x + 4Second term: (x + √10 - 2)² = [x + (√10 - 2)]² = x² + 2(√10 - 2)x + (√10 - 2)²Expanding (√10 - 2)² = (√10)² - 4√10 + 4 = 10 - 4√10 + 4 = 14 - 4√10So, expanding the second term:x² + 2(√10 - 2)x + 14 - 4√10Now, combine the two terms:First term: x² - 4x + 4Second term: x² + 2(√10 - 2)x + 14 - 4√10Adding them together:x² -4x +4 + x² + 2(√10 - 2)x +14 -4√10Combine like terms:2x² + (-4x + 2(√10 - 2)x) + (4 + 14 - 4√10)Simplify the x terms:-4x + 2(√10)x - 4x = (-4 - 4)x + 2√10 x = -8x + 2√10 x = x(-8 + 2√10)Constant terms:4 +14 = 18; 18 -4√10So total equation becomes:2x² + x(-8 + 2√10) + 18 -4√10 = 9Subtract 9 from both sides:2x² + (-8 + 2√10)x + 9 -4√10 = 0This is a quadratic equation in x. Let me write it as:2x² + ( -8 + 2√10 )x + (9 -4√10) = 0Let me divide all terms by 2 to simplify:x² + ( -4 + √10 )x + (9/2 - 2√10 ) = 0Hmm, solving this quadratic equation might be a bit messy, but let's try using the quadratic formula.Given ax² + bx + c = 0,x = [-b ± sqrt(b² -4ac)] / 2aHere, a = 1b = -4 + √10c = 9/2 - 2√10So, discriminant D = b² -4acLet's compute D:b² = (-4 + √10)^2 = 16 -8√10 +10 = 26 -8√104ac = 4*1*(9/2 -2√10) = 4*(9/2) -4*2√10 = 18 -8√10So, D = (26 -8√10) - (18 -8√10) = 26 -8√10 -18 +8√10 = 8So, discriminant D = 8Thus, x = [4 - √10 ± sqrt(8)] / 2sqrt(8) = 2*sqrt(2), so:x = [4 - √10 ± 2√2]/2 = [4 - √10]/2 ± √2Simplify:x = 2 - (√10)/2 ± √2So, the x-coordinates of the intersection points are 2 - (√10)/2 + √2 and 2 - (√10)/2 - √2.Now, to find the corresponding y-coordinates, since y = x + √10, we substitute:For x1 = 2 - (√10)/2 + √2:y1 = 2 - (√10)/2 + √2 + √10 = 2 + (√10)/2 + √2Similarly, for x2 = 2 - (√10)/2 - √2:y2 = 2 - (√10)/2 - √2 + √10 = 2 + (√10)/2 - √2Therefore, the two points of intersection are:Point A: (2 - (√10)/2 + √2, 2 + (√10)/2 + √2)Point B: (2 - (√10)/2 - √2, 2 + (√10)/2 - √2)Now, to find the distance between these two points, we can use the distance formula:Distance = sqrt[(x1 - x2)² + (y1 - y2)²]Compute x1 - x2:[2 - (√10)/2 + √2] - [2 - (√10)/2 - √2] = 2 - (√10)/2 + √2 -2 + (√10)/2 + √2 = 2√2Similarly, y1 - y2:[2 + (√10)/2 + √2] - [2 + (√10)/2 - √2] = 2 + (√10)/2 + √2 -2 - (√10)/2 + √2 = 2√2Therefore, the distance is sqrt[(2√2)^2 + (2√2)^2] = sqrt[8 + 8] = sqrt[16] = 4So, the chord length is indeed 4. That matches the previous method. So, that confirms that the chord length is 4. Therefore, the answer should be 4.But wait, let me just recap to make sure I didn't skip any steps or make any calculation errors.First method: distance from center to line is sqrt(5). Then chord length is 2*sqrt(r² - d²) = 2*sqrt(9 -5)=2*sqrt(4)=4. That seems straightforward.Second method: solving the system, found two points, computed their distance, got 4. So both methods agree. Therefore, the chord length is 4.But let me check if the distance was correctly calculated. The line is x - y + sqrt(10) =0. Center (2,2). So distance is |2 - 2 + sqrt(10)| / sqrt(1 +1) = |sqrt(10)| / sqrt(2) = sqrt(10)/sqrt(2) = sqrt(5) as sqrt(10/2) = sqrt(5). Then chord length 2*sqrt(9 -5)=4. Correct.All steps check out. Therefore, the answer is 4.Final Answerboxed{4}</think>To find the chord length cut by the line ( x - y + sqrt{10} = 0 ) from the circle ( x^2 + y^2 - 4x - 4y - 1 = 0 ):1. Convert the circle equation to standard form: [ x^2 + y^2 - 4x - 4y - 1 = 0 ] Completing the square for ( x ) and ( y ): [ (x - 2)^2 + (y - 2)^2 = 9 ] The circle has center ( (2, 2) ) and radius ( 3 ).2. Calculate the distance from the center to the line: The line equation is ( x - y + sqrt{10} = 0 ). Using the distance formula: [ text{Distance} = frac{|1 cdot 2 + (-1) cdot 2 + sqrt{10}|}{sqrt{1^2 + (-1)^2}} = frac{|sqrt{10}|}{sqrt{2}} = frac{sqrt{10}}{sqrt{2}} = sqrt{5} ]3. Compute the chord length: Using the formula for the chord length ( 2 sqrt{r^2 - d^2} ): [ text{Chord length} = 2 sqrt{3^2 - (sqrt{5})^2} = 2 sqrt{9 - 5} = 2 sqrt{4} = 4 ]4. Verification by finding intersection points: Solving the system of equations confirms the chord length calculation. The distance between the intersection points is found to be ( 4 ).Thus, the chord length is (boxed{4}).,<think>Okay, so I need to find the length of the chord that the line x - y + √10 = 0 cuts from the circle defined by the equation x² + y² - 4x - 4y - 1 = 0. Hmm, let's see. I remember there's a formula for the length of a chord in a circle when you know the distance from the center to the line. Let me try to recall that.First, maybe I should write the circle equation in standard form to find its center and radius. The given equation is x² + y² - 4x - 4y - 1 = 0. To convert this into standard form (which is (x - h)² + (y - k)² = r²), I need to complete the square for both the x and y terms.Starting with the x terms: x² - 4x. Completing the square here would involve taking half of -4, which is -2, squaring it to get 4, and adding it. Similarly for the y terms: y² - 4y. Half of -4 is also -2, squared is 4. So I need to add and subtract these values to complete the square.Let me rewrite the original equation with spaces for completing the square:x² - 4x + ___ + y² - 4y + ___ = 1 + ___ + ___Since we added 4 to both the x terms and the y terms, we need to add those to the right side as well. So:(x² - 4x + 4) + (y² - 4y + 4) = 1 + 4 + 4That simplifies to:(x - 2)² + (y - 2)² = 9So the circle M has center at (2, 2) and radius 3. Okay, that's straightforward.Now, the line given is x - y + √10 = 0. I need to find the length of the chord that this line cuts from the circle. As I mentioned, I think there's a formula involving the distance from the center to the line and then using the Pythagorean theorem to find half the chord length.The formula for the distance from a point (h, k) to the line Ax + By + C = 0 is |Ax + By + C| / sqrt(A² + B²). Let me confirm that. Yes, it's |A*h + B*k + C| / sqrt(A² + B²). So in this case, the line is x - y + √10 = 0. Comparing to Ax + By + C = 0, A is 1, B is -1, and C is √10. The center of the circle is (2, 2).Plugging into the distance formula: |1*2 + (-1)*2 + √10| / sqrt(1² + (-1)²) = |2 - 2 + √10| / sqrt(2) = |√10| / sqrt(2) = √10 / sqrt(2). Simplifying that, since √10 / sqrt(2) = sqrt(10/2) = sqrt(5). So the distance from the center to the line is √5.Now, the radius of the circle is 3. So if we imagine the line cutting the circle, the distance from the center to the line is √5, which is less than the radius 3, so the line does intersect the circle, forming a chord. The length of the chord can be found using the formula: 2 * sqrt(r² - d²), where r is the radius and d is the distance from the center to the line.Plugging in the values: 2 * sqrt(3² - (√5)²) = 2 * sqrt(9 - 5) = 2 * sqrt(4) = 2 * 2 = 4. So the length of the chord is 4. Wait, is that correct? Let me check again.Wait, the distance from the center to the line is √5, and the radius is 3. Then, the half-length of the chord is sqrt(r² - d²) = sqrt(9 - 5) = sqrt(4) = 2. So the full length is 2 * 2 = 4. Yeah, that seems right.But let me verify with another method, maybe by solving the system of equations to find the points of intersection and then compute the distance between them.First, let's parametrize the line. The line is x - y + √10 = 0, so we can write x = y - √10. Substitute this into the circle equation:(y - √10)² + y² - 4(y - √10) - 4y - 1 = 0.Expanding the terms:First, expand (y - √10)²: y² - 2y√10 + 10.Then, substitute back into the equation:(y² - 2y√10 + 10) + y² - 4y + 4√10 - 4y - 1 = 0.Combine like terms:y² - 2y√10 + 10 + y² - 4y + 4√10 - 4y - 1 = 0Combine the y² terms: 2y².Combine the y terms: (-2√10 - 4 - 4)y = (-2√10 - 8)y.Combine constants: 10 + 4√10 - 1 = 9 + 4√10.So the equation becomes:2y² + (-2√10 - 8)y + (9 + 4√10) = 0.This is a quadratic in terms of y. Let's write it as:2y² - (2√10 + 8)y + (9 + 4√10) = 0.To solve for y, we can use the quadratic formula. Let's compute discriminant D first.D = [-(2√10 + 8)]² - 4*2*(9 + 4√10)First compute [2√10 + 8]^2:= (2√10)^2 + 2*2√10*8 + 8^2= 4*10 + 32√10 + 64= 40 + 32√10 + 64= 104 + 32√10Now compute 4*2*(9 + 4√10):= 8*(9 + 4√10)= 72 + 32√10Thus, D = (104 + 32√10) - (72 + 32√10) = 104 - 72 = 32.So discriminant D = 32. Therefore, the solutions are:y = [(2√10 + 8) ± sqrt(32)] / (2*2)Simplify sqrt(32) = 4*sqrt(2). So:y = [2√10 + 8 ± 4√2] / 4Factor numerator by 2:y = 2[√10 + 4 ± 2√2] / 4 = [√10 + 4 ± 2√2]/2Therefore, the two y-coordinates are:y1 = [√10 + 4 + 2√2]/2 and y2 = [√10 + 4 - 2√2]/2Now, the x-coordinates can be found from x = y - √10. So:For y1: x1 = [√10 + 4 + 2√2]/2 - √10 = [√10 + 4 + 2√2 - 2√10]/2 = [4 + 2√2 - √10]/2Similarly, for y2: x2 = [√10 + 4 - 2√2]/2 - √10 = [√10 + 4 - 2√2 - 2√10]/2 = [4 - 2√2 - √10]/2So the two points of intersection are:P1: ( [4 + 2√2 - √10]/2 , [√10 + 4 + 2√2]/2 )P2: ( [4 - 2√2 - √10]/2 , [√10 + 4 - 2√2]/2 )Now, to find the distance between these two points. Let's denote the coordinates as:P1: ( (4 - √10 + 2√2)/2 , (4 + √10 + 2√2)/2 )P2: ( (4 - √10 - 2√2)/2 , (4 + √10 - 2√2)/2 )So, let's compute the differences in x and y coordinates:Δx = [ (4 - √10 + 2√2)/2 ] - [ (4 - √10 - 2√2)/2 ] = [4 - √10 + 2√2 - 4 + √10 + 2√2]/2 = (4√2)/2 = 2√2Similarly, Δy = [ (4 + √10 + 2√2)/2 ] - [ (4 + √10 - 2√2)/2 ] = [4 + √10 + 2√2 - 4 - √10 + 2√2]/2 = (4√2)/2 = 2√2Therefore, the distance between P1 and P2 is sqrt( (Δx)^2 + (Δy)^2 ) = sqrt( (2√2)^2 + (2√2)^2 ) = sqrt(8 + 8) = sqrt(16) = 4. So that's consistent with the previous result. So the chord length is indeed 4.Therefore, the answer should be 4, so boxed{4}.Final AnswerThe length of the chord is boxed{4}.</think>To find the length of the chord cut by the line ( x - y + sqrt{10} = 0 ) from the circle ( M: x^2 + y^2 - 4x - 4y - 1 = 0 ), we start by converting the circle equation to standard form. 1. Convert the circle equation to standard form: [ x^2 + y^2 - 4x - 4y - 1 = 0 ] Completing the square for ( x ) and ( y ): [ (x^2 - 4x + 4) + (y^2 - 4y + 4) = 1 + 4 + 4 ] Simplifying, we get: [ (x - 2)^2 + (y - 2)^2 = 9 ] The circle has center ((2, 2)) and radius (3).2. Calculate the distance from the center to the line: The line is ( x - y + sqrt{10} = 0 ). Using the distance formula from a point ((h, k)) to a line (Ax + By + C = 0): [ text{Distance} = frac{|1 cdot 2 + (-1) cdot 2 + sqrt{10}|}{sqrt{1^2 + (-1)^2}} = frac{|sqrt{10}|}{sqrt{2}} = frac{sqrt{10}}{sqrt{2}} = sqrt{5} ]3. Find the length of the chord: Using the formula for the length of a chord (2 sqrt{r^2 - d^2}) where (r) is the radius and (d) is the distance from the center to the line: [ text{Length} = 2 sqrt{3^2 - (sqrt{5})^2} = 2 sqrt{9 - 5} = 2 sqrt{4} = 4 ]4. Verification by solving the system: Substituting (x = y - sqrt{10}) into the circle equation and solving the quadratic equation confirms the points of intersection. Calculating the distance between these points also results in a chord length of (4).Thus, the length of the chord is (boxed{4}).]

🔑:From the equation of the circle x^2+y^2-4x-4y-1=0, we can find that the coordinates of the center of the circle are (2, 2), and the radius R=3.The distance d from the center of the circle to the line x-y+sqrt{10}=0 is frac{sqrt{10}}{sqrt{2}}=sqrt{5}.By forming a right-angled triangle with the half-chord length, the distance from the chord to the center, and the radius, and applying the Pythagorean theorem, we get:l=2sqrt{9-5}=4.Therefore, the answer is boxed{4}.From the given equations of the line and the circle, we can derive the general equation of the line, the coordinates of the center of the circle, and the radius. By forming a right-angled triangle with the half-chord length, the distance from the chord to the center, and the radius, and applying the Pythagorean theorem, we can find the answer.This problem tests the application of the equations of lines and circles, where the problem of the chord length intersected by a line and a circle is often solved by forming a right-angled triangle with the half-chord length, the distance from the chord to the center, and the radius, and applying the Pythagorean theorem.

🧠:[<think>Okay, so I need to solve this problem about the sequence {a_n} with the given recurrence relation and initial terms. Part (1) asks me to prove that the sequence {b_n}, defined as b_n = a_n / (a_{n+1} - a_n), is an arithmetic sequence. Part (2) is about finding the sum T_n of the first n terms of {c_n}, where c_n = 1 / (a_n a_{n+1}).Starting with part (1). Let me first write down what's given. The recurrence relation is (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n (a_{n+2} - a_{n+1}) for all natural numbers n. Also, a_1 = 1 and a_2 = 2. First, maybe I should compute the first few terms of {a_n} to see if I can spot a pattern or get some intuition. Let's try computing a_3, a_4, etc., using the recurrence.Given a_1 = 1 and a_2 = 2. Let's plug in n = 1 into the recurrence:For n = 1:Left-hand side: (a_2 - a_1)(2a_2 - a_3) = (2 - 1)(4 - a_3) = 1*(4 - a_3) = 4 - a_3Right-hand side: a_1(a_3 - a_2) = 1*(a_3 - 2) = a_3 - 2Set them equal:4 - a_3 = a_3 - 2Solving for a_3:4 + 2 = a_3 + a_3 => 6 = 2a_3 => a_3 = 3Wait, that's a problem. Wait, a_3 is computed as 3? Let me check the calculations again.Left-hand side: (a_2 - a_1) = 2 - 1 = 1. Then 2a_2 - a_3 = 4 - a_3. Multiply those: 1*(4 - a_3) = 4 - a_3.Right-hand side: a_1*(a_3 - a_2) = 1*(a_3 - 2) = a_3 - 2.Setting equal: 4 - a_3 = a_3 - 2. Then 4 + 2 = 2a_3 => 6 = 2a_3 => a_3 = 3. Hmm, okay. So a_3 is 3.Wait, but if a_1 = 1, a_2 = 2, a_3 = 3, maybe the sequence is just a_n = n? Let's check n=2 next to see.For n = 2:Left-hand side: (a_3 - a_2)(2a_3 - a_4) = (3 - 2)(6 - a_4) = 1*(6 - a_4) = 6 - a_4Right-hand side: a_2*(a_4 - a_3) = 2*(a_4 - 3) = 2a_4 - 6Set equal: 6 - a_4 = 2a_4 - 6Solving: 6 + 6 = 2a_4 + a_4 => 12 = 3a_4 => a_4 = 4Hmm, so a_4 = 4? Then a_5 would be 5? Let me test n=3.For n = 3:Left-hand side: (a_4 - a_3)(2a_4 - a_5) = (4 - 3)(8 - a_5) = 1*(8 - a_5) = 8 - a_5Right-hand side: a_3*(a_5 - a_4) = 3*(a_5 - 4) = 3a_5 - 12Set equal: 8 - a_5 = 3a_5 - 12Solving: 8 + 12 = 3a_5 + a_5 => 20 = 4a_5 => a_5 = 5So, indeed, the sequence seems to be a_n = n. But wait, if that's the case, then {a_n} is just 1, 2, 3, 4, 5,... So each term is n. Then, let's check if this satisfies the given recurrence relation for general n.Assume a_n = n. Then, check if (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})Compute LHS: ( (n+1) - n )( 2(n+1) - (n+2) ) = (1)(2n + 2 - n - 2) = (1)(n) = nRHS: a_n (a_{n+2} - a_{n+1}) = n*( (n+2) - (n+1) ) = n*(1) = nSo LHS = RHS = n. Therefore, a_n = n satisfies the recurrence. And since the initial terms are 1 and 2, which match a_1 = 1 and a_2 = 2, the sequence is indeed a_n = n for all n.But wait, that seems too straightforward. If the sequence is just a_n = n, then part (1) would be easy. Let's check.Given that a_n = n, then a_{n+1} - a_n = (n+1) - n = 1. Therefore, b_n = a_n / (a_{n+1} - a_n) = n / 1 = n. Therefore, the sequence {b_n} is 1, 2, 3, 4,... which is an arithmetic sequence with common difference 1. Hence, part (1) is proven.But maybe the problem is designed to not assume that a_n = n, but to derive it. But since with the given initial terms and recurrence, the sequence is forced to be a_n = n, then that's the case.Wait, but the problem could be more complex. Let me double-check. Suppose I didn't notice that a_n = n, how would I approach part (1)? Maybe I need to manipulate the recurrence relation given to find a relation for b_n.Given that b_n = a_n / (a_{n+1} - a_n). Let's try to express the given recurrence in terms of b_n.The given recurrence is (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})Let me denote d_n = a_{n+1} - a_n. Then, the given equation becomes:d_n (2a_{n+1} - a_{n+2}) = a_n (a_{n+2} - a_{n+1})Note that a_{n+2} - a_{n+1} = d_{n+1}Also, 2a_{n+1} - a_{n+2} = 2a_{n+1} - (a_{n+1} + d_{n+1}) ) = a_{n+1} - d_{n+1}Therefore, substituting into the equation:d_n (a_{n+1} - d_{n+1}) = a_n d_{n+1}Let's expand the left side:d_n a_{n+1} - d_n d_{n+1} = a_n d_{n+1}Bring all terms to one side:d_n a_{n+1} = a_n d_{n+1} + d_n d_{n+1}Factor d_{n+1} on the right:d_n a_{n+1} = d_{n+1} (a_n + d_n)But a_n + d_n = a_n + (a_{n+1} - a_n) = a_{n+1}Therefore, the equation becomes:d_n a_{n+1} = d_{n+1} a_{n+1}Assuming a_{n+1} ≠ 0, we can divide both sides by a_{n+1}:d_n = d_{n+1}Wait, this suggests that d_n is constant for all n. If d_n = d_{n+1} for all n, then the difference between consecutive terms is constant, i.e., the sequence {a_n} is arithmetic. But in our case, we found that a_n = n, which is an arithmetic sequence with common difference 1, so d_n = 1 for all n. Hence, this is consistent.But wait, how does this relate to b_n? Remember that b_n = a_n / d_n. If d_n is constant, then b_n = a_n / d. Since a_n is arithmetic (linear), then b_n would also be arithmetic. For example, if a_n = a_1 + (n-1)d_a, then b_n = (a_1 + (n-1)d_a)/d_d, which is linear in n, hence arithmetic.But in our specific case, a_n = n, so d_n = 1, hence b_n = n / 1 = n, which is arithmetic with common difference 1.But the problem says to prove that {b_n} is an arithmetic sequence. So perhaps in general, even if {a_n} is not linear, but given the recurrence, {b_n} is arithmetic.Wait, but in our case, {a_n} turned out to be linear. So perhaps the key is to show that d_n is constant, hence leading to {a_n} being arithmetic, which in turn makes {b_n} arithmetic as well. But maybe the problem expects a more general approach without assuming {a_n} is linear.Wait, but according to the recurrence, we have d_n = d_{n+1}, so all d_n are equal, so d_n is a constant. Therefore, {a_n} is an arithmetic sequence, hence a_n = a_1 + (n-1)d. Given a_1 = 1, a_2 = 2, so d = 1, hence a_n = 1 + (n-1)*1 = n. Therefore, indeed, {a_n} is arithmetic with d = 1.Therefore, {b_n} = a_n / d_n = n / 1 = n, which is arithmetic with common difference 1.But let me check this again. If d_n is constant, then b_n = a_n / d. Since a_n is arithmetic, which is a linear function, then dividing by a constant d would still give a linear function, hence arithmetic. Therefore, {b_n} is arithmetic. So that's the proof.Therefore, part (1) can be proven by showing that the differences d_n = a_{n+1} - a_n are constant, hence {a_n} is arithmetic, hence {b_n} is arithmetic.Alternatively, we can directly find a relation between b_n and b_{n+1} and show that the difference b_{n+1} - b_n is constant.Given that b_n = a_n / (a_{n+1} - a_n). Since we already found that d_n = a_{n+1} - a_n is constant, say d, then b_n = a_n / d. Since a_n is arithmetic, a_n = a_1 + (n - 1)d. Therefore, b_n = [a_1 + (n - 1)d]/d = (a_1)/d + (n - 1). Therefore, b_n is an arithmetic sequence with first term (a_1)/d and common difference 1. In our case, a_1 = 1 and d = 1, so b_n = 1 + (n - 1) = n, which is arithmetic with common difference 1.Therefore, the general proof would be: Since the recurrence relation implies that d_n = d_{n+1} for all n, the differences d_n are constant. Therefore, {a_n} is an arithmetic sequence with common difference d = a_{n+1} - a_n = 1 (since a_2 - a_1 = 1). Therefore, a_n = 1 + (n - 1)*1 = n. Then, b_n = a_n / (a_{n+1} - a_n) = n / 1 = n, which is an arithmetic sequence with common difference 1. Therefore, the conclusion holds.Alternatively, even without knowing that {a_n} is arithmetic, since we have established that d_n is constant, then b_n = a_n / d is linear in n, hence arithmetic.Therefore, part (1) is proven.Moving on to part (2). We need to find the sum T_n of the first n terms of {c_n}, where c_n = 1 / (a_n a_{n+1}).Given that a_n = n, as established, then c_n = 1 / [n(n + 1)]. Therefore, the sum T_n is the sum from k = 1 to n of 1 / [k(k + 1)].This is a telescoping series. Recall that 1 / [k(k + 1)] = 1/k - 1/(k + 1). Therefore, when we sum from k = 1 to n, most terms cancel out:Sum_{k=1}^n [1/k - 1/(k+1)] = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1)) = 1 - 1/(n + 1) = n / (n + 1).Therefore, T_n = n / (n + 1).But let me verify this with the actual terms. For n = 1, c_1 = 1/(1*2) = 1/2, so T_1 = 1/2 = 1/(1 + 1). For n = 2, c_1 + c_2 = 1/2 + 1/(2*3) = 1/2 + 1/6 = 2/3 = 2/(2 + 1). For n = 3, sum is 1/2 + 1/6 + 1/(3*4) = 1/2 + 1/6 + 1/12 = (6 + 2 + 1)/12 = 9/12 = 3/4 = 3/(3 + 1). So yes, the formula holds. Therefore, T_n = n / (n + 1).But let me confirm this through the problem's lens. Since the problem gives a specific definition for c_n and asks for the sum, even if we didn't compute the terms, the telescoping approach is standard for such fractions.Therefore, the answer is T_n = n / (n + 1).So, summarizing:1) Proved that {b_n} is arithmetic because {a_n} is arithmetic with common difference 1, hence b_n = n, which is arithmetic.2) Found that T_n = n/(n + 1) using telescoping series.Final Answer(1) boxed{b_n} is an arithmetic sequence.(2) The sum of the first ( n ) terms ( T_n ) is boxed{dfrac{n}{n + 1}}.</think>Given the sequence ({a_n}) with the recurrence relation ((a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})) and initial terms (a_1 = 1) and (a_2 = 2):1. Proving that ({b_n}) is an arithmetic sequence: - Compute the first few terms of ({a_n}) using the recurrence relation: - For (n = 1): (a_3 = 3) - For (n = 2): (a_4 = 4) - For (n = 3): (a_5 = 5) - Observing the pattern, we find (a_n = n). - Verify that (a_n = n) satisfies the recurrence relation, confirming it is correct. - Define (d_n = a_{n+1} - a_n). Since (a_n = n), (d_n = 1) for all (n), showing ({a_n}) is arithmetic with common difference 1. - Define (b_n = frac{a_n}{a_{n+1} - a_n} = frac{n}{1} = n), which is an arithmetic sequence with common difference 1.2. Finding the sum (T_n) of the first (n) terms of ({c_n}): - Given (c_n = frac{1}{a_n a_{n+1}}) and (a_n = n), we have (c_n = frac{1}{n(n+1)}). - Use the telescoping series property: [ sum_{k=1}^n frac{1}{k(k+1)} = sum_{k=1}^n left( frac{1}{k} - frac{1}{k+1} right) ] - This simplifies to: [ left(1 - frac{1}{2}right) + left(frac{1}{2} - frac{1}{3}right) + cdots + left(frac{1}{n} - frac{1}{n+1}right) = 1 - frac{1}{n+1} = frac{n}{n+1} ] Final Answer(1) (boxed{b_n}) is an arithmetic sequence.(2) The sum of the first (n) terms (T_n) is (boxed{dfrac{n}{n + 1}}).,<think>Okay, let's try to tackle this problem step by step. So, we have a sequence {a_n} defined by the recurrence relation (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1}) for all natural numbers n, with initial conditions a₁ = 1 and a₂ = 2. The problem has two parts: first, to prove that the sequence {b_n} defined by b_n = a_n / (a_{n+1} - a_n) is an arithmetic sequence, and second, to find the sum T_n of the first n terms of {c_n} where c_n = 1/(a_n a_{n+1}).Starting with part (1). We need to show that {b_n} is arithmetic, which means that the difference between consecutive terms should be constant. An arithmetic sequence has the form b_n = b₁ + (n - 1)d for some common difference d. So, if we can show that b_{n+1} - b_n is a constant, that would prove it's arithmetic.Given that b_n = a_n / (a_{n+1} - a_n). Let me try to find an expression for b_{n+1} in terms of a_{n+1} and a_{n+2}, and then see if we can relate it to b_n using the given recurrence relation.First, write down b_{n+1}:b_{n+1} = a_{n+1} / (a_{n+2} - a_{n+1})So, if we can express a_{n+2} - a_{n+1} in terms of a_n and a_{n+1}, that might help. The recurrence relation given is:(a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})Let me rearrange this equation to solve for a_{n+2}. Let's denote for simplicity:Let’s set x = a_{n+1} - a_n. Then the left side becomes x*(2a_{n+1} - a_{n+2}) and the right side is a_n*(a_{n+2} - a_{n+1}).Wait, perhaps expanding both sides would be better. Let's expand the left-hand side (LHS):LHS = (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = 2a_{n+1}(a_{n+1} - a_n) - a_{n+2}(a_{n+1} - a_n)Right-hand side (RHS):RHS = a_n(a_{n+2} - a_{n+1}) = a_n a_{n+2} - a_n a_{n+1}So, setting LHS = RHS:2a_{n+1}(a_{n+1} - a_n) - a_{n+2}(a_{n+1} - a_n) = a_n a_{n+2} - a_n a_{n+1}Let me collect terms involving a_{n+2} on one side:- a_{n+2}(a_{n+1} - a_n) - a_n a_{n+2} = -2a_{n+1}(a_{n+1} - a_n) - a_n a_{n+1}Factor a_{n+2} on the left:a_{n+2}[ - (a_{n+1} - a_n) - a_n ] = -2a_{n+1}(a_{n+1} - a_n) - a_n a_{n+1}Simplify the coefficients:Left side coefficient:- (a_{n+1} - a_n) - a_n = -a_{n+1} + a_n - a_n = -a_{n+1}Right side:-2a_{n+1}(a_{n+1} - a_n) - a_n a_{n+1} = -2a_{n+1}^2 + 2a_n a_{n+1} - a_n a_{n+1} = -2a_{n+1}^2 + a_n a_{n+1}So, we have:- a_{n+1} a_{n+2} = -2a_{n+1}^2 + a_n a_{n+1}Divide both sides by -a_{n+1} (assuming a_{n+1} ≠ 0, which seems reasonable as a₁=1, a₂=2, and the sequence seems to be increasing, but we can check later):a_{n+2} = 2a_{n+1} - a_nWait, so the recurrence simplifies to a_{n+2} = 2a_{n+1} - a_n. That's a linear recurrence relation! So, the original complicated-looking recurrence actually reduces to a linear one with constant coefficients. That seems much more manageable.So, the recurrence is a_{n+2} = 2a_{n+1} - a_n. Let me check this with the initial terms. Given a₁ = 1, a₂ = 2. Let's compute a₃:Using the recurrence for n=1:a₃ = 2a₂ - a₁ = 2*2 - 1 = 4 - 1 = 3.Similarly, a₄ = 2a₃ - a₂ = 2*3 - 2 = 6 - 2 = 4.a₅ = 2a₄ - a₃ = 2*4 - 3 = 8 - 3 = 5.Wait a second, this seems like the sequence is a_n = n. Because a₁=1, a₂=2, a₃=3, a₄=4, a₅=5, etc. So maybe a_n = n? Let's check if this satisfies the original recurrence.Original recurrence: (a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})If a_n = n, then a_{n+1} - a_n = 1, 2a_{n+1} - a_{n+2} = 2(n+1) - (n+2) = 2n + 2 - n - 2 = n. On the left side: 1 * n = n.On the right side: a_n(a_{n+2} - a_{n+1}) = n*( (n+2) - (n+1) ) = n*(1) = n. So yes, equality holds. Therefore, the solution a_n = n satisfies both the recurrence and the initial conditions. Therefore, the sequence {a_n} is simply a_n = n.Wait, that seems too straightforward. If a_n = n, then the problem becomes trivial. Let me verify with n=1:Given a₁=1, a₂=2, so a₃=3. Then check the original equation for n=1:(a₂ - a₁)(2a₂ - a₃) = (2 - 1)(4 - 3) = 1*1 = 1a₁(a₃ - a₂) = 1*(3 - 2) = 1*1 = 1. So it works.Similarly, n=2:(a₃ - a₂)(2a₃ - a₄) = (3 - 2)(6 - 4) = 1*2 = 2a₂(a₄ - a₃) = 2*(4 - 3) = 2*1 = 2. Also works.So indeed, the sequence is a_n = n. Therefore, the problem's sequence {a_n} is just the natural numbers. Therefore, part (1) becomes straightforward.Given a_n = n, then a_{n+1} - a_n = (n + 1) - n = 1. Therefore, b_n = a_n / (a_{n+1} - a_n) = n / 1 = n. Therefore, {b_n} is the sequence 1, 2, 3, 4,... which is clearly an arithmetic sequence with common difference 1. Therefore, part (1) is proven.But wait, perhaps the problem expects us to derive this without assuming a_n = n? Because maybe in the original problem, even though we can see that a_n = n satisfies the recurrence, perhaps we need to show that {b_n} is arithmetic regardless of the specific solution? Wait, no, since the problem gives specific initial conditions a₁=1 and a₂=2, which leads to a_n = n. Therefore, in this specific case, the sequence is arithmetic. But maybe in general, even if the recurrence a_{n+2} = 2a_{n+1} - a_n is a linear recurrence with characteristic equation r² - 2r + 1 = 0, which has a double root at r=1. Therefore, the general solution is a_n = (A + Bn)(1)^n = A + Bn. Given the initial conditions a₁=1, a₂=2:For n=1: A + B(1) = 1For n=2: A + B(2) = 2Subtracting first equation from second: B = 1. Then A = 1 - B = 0. Therefore, a_n = 0 + 1*n = n. Therefore, indeed, the solution is a_n = n. So, the problem is specifically about this sequence. Therefore, {b_n} is n, which is arithmetic.But the problem says "prove that {b_n} is an arithmetic sequence". So, maybe even if we didn't solve for a_n, we can use the recurrence a_{n+2} = 2a_{n+1} - a_n to show that b_n is arithmetic. Let's see.Given that a_{n+2} = 2a_{n+1} - a_n. Then, let's express b_n = a_n / (a_{n+1} - a_n). Let's find a relation between b_{n+1} and b_n.First, note that from the recurrence, a_{n+2} - a_{n+1} = (2a_{n+1} - a_n) - a_{n+1} = a_{n+1} - a_n. Therefore, a_{n+2} - a_{n+1} = a_{n+1} - a_n. So the differences between consecutive terms are constant? Wait, if a_{n+1} - a_n is constant, then the sequence {a_n} is arithmetic. Indeed, since a_{n+1} - a_n = d (constant), then a_n = a₁ + (n -1)d. But in our case, a_{n+2} - a_{n+1} = a_{n+1} - a_n, which implies that the difference is the same for all n. Therefore, {a_n} is an arithmetic sequence. However, given the initial terms a₁=1, a₂=2, the common difference d=1, so indeed a_n = 1 + (n -1)*1 = n. So that's another way to see it.But wait, hold on. If a_{n+1} - a_n is constant, then the sequence is arithmetic. Therefore, the recurrence a_{n+2} = 2a_{n+1} - a_n is the standard recurrence for an arithmetic sequence. Because in an arithmetic sequence, each term is the previous term plus d, so a_{n+2} = a_{n+1} + d = a_{n} + 2d, which can be written as a_{n+2} = 2a_{n+1} - a_n. Therefore, yes, the recurrence is characteristic of an arithmetic sequence. Therefore, given that, {a_n} is arithmetic, hence a_n = n.But since the problem is structured to first prove that {b_n} is arithmetic, maybe they expect us to do it without assuming {a_n} is arithmetic? Let me see.Given b_n = a_n / (a_{n+1} - a_n). Let’s compute b_{n+1} - b_n and show it's constant.From the recurrence a_{n+2} = 2a_{n+1} - a_n, which we derived earlier. Let's compute a_{n+1} - a_n. Let’s denote d_n = a_{n+1} - a_n. Then from the recurrence, we have:a_{n+2} - a_{n+1} = (2a_{n+1} - a_n) - a_{n+1} = a_{n+1} - a_n = d_nTherefore, d_{n+1} = d_n. So the difference d_n is constant. Therefore, d_n = d for all n. Therefore, {a_n} is arithmetic with common difference d. Since a₁=1, a₂=2, d=1, so a_n = 1 + (n -1)*1 = n.But since we have d_n = d_{n+1}, then d is constant, so a_{n+1} - a_n = d. Therefore, b_n = a_n / d. Since d=1, b_n = a_n. But since a_n is arithmetic, {b_n} is also arithmetic. Therefore, that's another way to see it.But the problem probably wants us to derive this step by step, using the given recurrence. So let's formalize it.We have to prove {b_n} is arithmetic. Let's find a relation between b_{n+1} and b_n.From the definition:b_n = a_n / (a_{n+1} - a_n)Similarly,b_{n+1} = a_{n+1} / (a_{n+2} - a_{n+1})From the recurrence, we have:a_{n+2} = 2a_{n+1} - a_nTherefore,a_{n+2} - a_{n+1} = a_{n+1} - a_nSo, the denominator in b_{n+1} is the same as the denominator in b_n:a_{n+2} - a_{n+1} = a_{n+1} - a_n = d (constant)Wait, so if a_{n+1} - a_n is constant, then d is 1, as we saw. Therefore, a_{n+1} - a_n = 1 for all n. Therefore, b_n = a_n / 1 = a_n = n. Hence, {b_n} is 1, 2, 3, ..., which is arithmetic with common difference 1.Alternatively, if we didn't realize that the difference is constant, but from the recurrence a_{n+2} - a_{n+1} = a_{n+1} - a_n, this implies that the difference is constant. Therefore, d_n = a_{n+1} - a_n is constant for all n, hence {a_n} is arithmetic. Therefore, {b_n} = a_n / d = (a₁ + (n - 1)d)/d = (1 + (n -1)*1)/1 = n, which is arithmetic.Therefore, part (1) is proven.Moving to part (2): Let c_n = 1/(a_n a_{n+1}), find the sum T_n of the first n terms of {c_n}.Given that a_n = n, so c_n = 1/(n(n + 1)). The sum T_n is the sum from k=1 to n of 1/(k(k + 1)).This is a telescoping series. Recall that 1/(k(k + 1)) = 1/k - 1/(k + 1). Therefore, the sum T_n = Σ_{k=1}^n [1/k - 1/(k + 1)].When we expand this sum, most terms cancel:(1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1))All the intermediate terms cancel, leaving T_n = 1 - 1/(n + 1) = n/(n + 1).Therefore, the sum of the first n terms of {c_n} is n/(n + 1).But let me verify this with the first few terms. For n=1: c₁ = 1/(1*2) = 1/2, so T₁ = 1/2. According to the formula, 1/(1 + 1) = 1/2. Correct.For n=2: c₁ + c₂ = 1/2 + 1/(2*3) = 1/2 + 1/6 = 2/3. The formula gives 2/(2 + 1) = 2/3. Correct.For n=3: 1/2 + 1/6 + 1/12 = (6 + 2 + 1)/12 = 9/12 = 3/4. The formula gives 3/(3 + 1) = 3/4. Correct. So the formula seems to hold.Therefore, T_n = n/(n + 1).But just to be thorough, let's go through the steps again. Given c_n = 1/(a_n a_{n+1}) = 1/(n(n + 1)). The sum T_n = Σ_{k=1}^n 1/(k(k + 1)).Partial fraction decomposition gives 1/(k(k + 1)) = 1/k - 1/(k + 1). Therefore:T_n = Σ_{k=1}^n [1/k - 1/(k + 1)] = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1)) = 1 - 1/(n + 1) = n/(n + 1)Yes, that's correct.Therefore, the answer to part (2) is T_n = n/(n + 1).So, summarizing:1) Proved that {b_n} is arithmetic by showing that the differences a_{n+1} - a_n are constant, leading to b_n = n, which is arithmetic.2) Found that the sum T_n is a telescoping series resulting in T_n = n/(n + 1).Final Answer(1) boxed{b_{n} = n}(2) boxed{T_{n} = dfrac{n}{n + 1}}</think>Given the sequence ({a_n}) defined by the recurrence relation ((a_{n+1} - a_n)(2a_{n+1} - a_{n+2}) = a_n(a_{n+2} - a_{n+1})) with initial conditions (a_1 = 1) and (a_2 = 2):1. Proving that ({b_n}) is an arithmetic sequence: - We start by simplifying the given recurrence relation. Through algebraic manipulation, we find that the recurrence simplifies to (a_{n+2} = 2a_{n+1} - a_n). This indicates that the sequence ({a_n}) is an arithmetic sequence with a common difference of 1, leading to (a_n = n). - The sequence (b_n) is defined as (b_n = frac{a_n}{a_{n+1} - a_n}). Since (a_{n+1} - a_n = 1) for all (n), we have (b_n = frac{a_n}{1} = a_n = n). This shows that ({b_n}) is an arithmetic sequence with a common difference of 1.2. Finding the sum of the first (n) terms of ({c_n}): - The sequence (c_n) is defined as (c_n = frac{1}{a_n a_{n+1}}). Given (a_n = n), we have (c_n = frac{1}{n(n + 1)}). - Using partial fraction decomposition, we express (c_n) as (frac{1}{n} - frac{1}{n + 1}). The sum (T_n) of the first (n) terms is a telescoping series: [ T_n = sum_{k=1}^n left(frac{1}{k} - frac{1}{k + 1}right) = 1 - frac{1}{n + 1} = frac{n}{n + 1} ] Final Answer(1) (boxed{b_{n} = n})(2) (boxed{T_{n} = dfrac{n}{n + 1}})]

🔑: Solution:# Part (1):Given the sequence {a_{n}} satisfies (a_{n+1}-a_{n})(2a_{n+1}-a_{n+2})=a_{n}(a_{n+2}-a_{n+1}) for nin N^{*}, and a_{1}=1, a_{2}=2.Step 1: Rewrite the given equation by distributing the terms:[(a_{n+1}-a_{n})(a_{n+1}+a_{n+1}-a_{n+2})=a_{n}(a_{n+2}-a_{n+1})]Step 2: Simplify the left-hand side of the equation:[a_{n+1}(a_{n+1}-a_{n})-(a_{n+1}-a_{n})(a_{n+2}-a_{n+1})=a_{n}(a_{n+2}-a_{n+1})]Step 3: Cancel out (a_{n+2}-a_{n+1}) on both sides:[a_{n+1}(a_{n+1}-a_{n})=(a_{n+2}-a_{n+1})a_{n+1}]Step 4: Since a_{n+1} neq 0, we can deduce that:[a_{n+1}-a_{n}=a_{n+2}-a_{n+1}]This implies 2a_{n+1}=a_{n}+a_{n+2}, showing that {a_{n}} is an arithmetic sequence with a common difference d.Step 5: Determine the difference between consecutive terms of {b_{n}}:[b_{n+1}-b_{n}=frac{a_{n+1}}{a_{n+2}-a_{n+1}}-frac{a_{n}}{a_{n+1}-a_{n}}=frac{a_{n+1}}{d}-frac{a_{n}}{d}=frac{d}{d}=1]Since b_{n+1}-b_{n}=1 is constant, {b_{n}} is an arithmetic sequence.Therefore, we have proven that {b_{n}} is an arithmetic sequence.# Part (2):Given that {a_{n}} is an arithmetic sequence with a_{1}=1 and a_{2}=2, the common difference d=1.Step 1: Express a_{n} in terms of n:[a_{n}=1+(n-1)times 1=n]Step 2: Calculate c_{n}:[c_{n}=frac{1}{a_{n}a_{n+1}}=frac{1}{n(n+1)}=frac{1}{n}-frac{1}{n+1}]Step 3: Find the sum T_{n} of the first n terms of {c_{n}}:[T_{n}=(1-frac{1}{2})+(frac{1}{2}-frac{1}{3})+ldots +(frac{1}{n}-frac{1}{n+1})=1-frac{1}{n+1}=frac{n}{n+1}]Hence, the sum of the first n terms of {c_{n}} is boxed{frac{n}{n+1}}.

🧠:[<think>Okay, let's see. The problem is about two functions, f(x) = x² - a x and g(x) = b + a ln(x - 1). We need to find the range of the real number b such that there's no common point between the graphs of y = f(x) and y = g(x). The answer choices are given, so maybe after solving, we can check which one it matches.First, the problem states that a is a real number greater than or equal to 1. So a ≥ 1. We need to find the range of b for which the two graphs don't intersect. That means the equation f(x) = g(x) has no real solutions. So, x² - a x = b + a ln(x - 1). Let me rearrange this equation to x² - a x - a ln(x - 1) - b = 0. We need this equation to have no real solutions. Therefore, the function h(x) = x² - a x - a ln(x - 1) - b should never be zero for any x in the domain.But wait, what's the domain here? The function g(x) has ln(x - 1), so x - 1 > 0, so x > 1. Therefore, the domain we're considering is x > 1. So we need to make sure that h(x) ≠ 0 for all x > 1. So we need to ensure that h(x) is always positive or always negative for x > 1. But since f(x) is a quadratic and g(x) is a logarithmic function, their behavior as x approaches infinity might tell us something. Let's check the limits as x approaches infinity.As x → ∞, x² - a x tends to infinity because the x² term dominates. On the other hand, a ln(x - 1) grows much slower, so g(x) tends to infinity as x → ∞ but much slower. Therefore, h(x) = f(x) - g(x) tends to infinity as x → ∞. So h(x) goes to infinity. Therefore, if h(x) is always positive for x > 1, then the graphs don't intersect. If h(x) ever dips below zero somewhere, then there would be an intersection. So we need to ensure that h(x) is always positive. That is, the minimum value of h(x) on x > 1 is greater than zero. Therefore, we can find the minimum of h(x) and set that minimum to be greater than zero. Then solve for b in terms of a. Then, considering a ≥ 1, find the range of b.So the plan is: find the critical points of h(x) by taking its derivative, set derivative equal to zero, find the x that minimizes h(x), plug that x back into h(x), set h(x) > 0, solve for b. Then, since a is ≥ 1, find the range of b accordingly.Let me start by finding the derivative of h(x). h(x) = x² - a x - a ln(x - 1) - b. So h'(x) = 2x - a - a/(x - 1). Let's set h'(x) = 0.2x - a - a/(x - 1) = 0.Let me rewrite this equation:2x = a + a/(x - 1)Multiply both sides by (x - 1):2x(x - 1) = a(x - 1) + aExpand the left side: 2x² - 2xRight side: a(x - 1) + a = a x - a + a = a xSo equation becomes 2x² - 2x = a xBring all terms to left side: 2x² - 2x - a x = 0Factor x: 2x² - (2 + a)x = 0x(2x - (2 + a)) = 0Solutions: x = 0 or 2x - (2 + a) = 0 => x = (2 + a)/2But the domain is x > 1, so x = 0 is rejected. So the critical point is at x = (2 + a)/2. Let's check if this is greater than 1.Since a ≥ 1, then (2 + a)/2 ≥ (2 + 1)/2 = 3/2 = 1.5 > 1. So yes, the critical point is in the domain.So the minimum of h(x) occurs at x = (2 + a)/2. Let's compute h at this point.First, compute h(x) = x² - a x - a ln(x - 1) - bLet me plug in x = (2 + a)/2.First, compute each term:x = (2 + a)/2x² = [(2 + a)/2]^2 = (4 + 4a + a²)/4a x = a*(2 + a)/2 = (2a + a²)/2ln(x - 1) = ln[(2 + a)/2 - 1] = ln[(2 + a - 2)/2] = ln(a/2) = ln a - ln 2Therefore, h(x) = x² - a x - a ln(x - 1) - bSubstitute:= (4 + 4a + a²)/4 - (2a + a²)/2 - a(ln a - ln 2) - bSimplify term by term:First term: (4 + 4a + a²)/4Second term: -(2a + a²)/2 = - (2a + a²)/2Third term: -a(ln a - ln 2) = -a ln a + a ln 2Fourth term: -bSo combine first and second term:(4 + 4a + a²)/4 - (2a + a²)/2 = Let's convert to quarters:= (4 + 4a + a²)/4 - (4a + 2a²)/4= [4 + 4a + a² - 4a - 2a²]/4= (4 - a²)/4So combining those gives (4 - a²)/4Then third term: -a ln a + a ln 2Fourth term: -bSo total h(x) = (4 - a²)/4 - a ln a + a ln 2 - bTherefore, h(x) at critical point is:(4 - a²)/4 - a ln a + a ln 2 - bWe need h(x) > 0 for the graphs not to intersect. So:(4 - a²)/4 - a ln a + a ln 2 - b > 0Solving for b:-b > - (4 - a²)/4 + a ln a - a ln 2Multiply both sides by -1 (inequality reverses):b < (4 - a²)/4 - a ln a + a ln 2So the upper bound for b is (4 - a²)/4 - a ln a + a ln 2Therefore, the range of b is all real numbers less than that expression. However, we need to find the range of b such that there exists a real number a ≥ 1 where this inequality holds. So for a given a ≥ 1, the maximum value of b is that expression. So to have such a exist, b must be less than the maximum of this expression over a ≥ 1. Wait, but actually, the problem states "there exists a real number a (a ≥ 1) such that the graph of y = f(x) has no common points with the graph of y = g(x)". So for a given b, we need there to exist at least one a ≥ 1 such that the inequality b < (4 - a²)/4 - a ln a + a ln 2 holds. So the range of b is all real numbers less than the maximum of (4 - a²)/4 - a ln a + a ln 2 over a ≥ 1.Therefore, we need to find the maximum value of the function k(a) = (4 - a²)/4 - a ln a + a ln 2 for a ≥ 1. Then the range of b is all real numbers less than this maximum. Therefore, the answer should be the interval (-∞, max k(a)).So now, we need to find the maximum of k(a) for a ≥ 1.Let me compute k(a):k(a) = (4 - a²)/4 - a ln a + a ln 2Simplify:= 1 - (a²)/4 - a ln a + a ln 2= 1 - (a²)/4 + a(ln 2 - ln a)= 1 - (a²)/4 + a ln(2/a)So we need to maximize k(a) over a ≥ 1. Let's take the derivative of k(a) with respect to a:k'(a) = d/da [1 - (a²)/4 + a ln(2/a)]Differentiate term by term:d/da [1] = 0d/da [-(a²)/4] = - (2a)/4 = -a/2d/da [a ln(2/a)] = Use product rule: first derivative of a times ln(2/a) + a times derivative of ln(2/a)Derivative of a is 1, so first term: ln(2/a)Second term: a * derivative of ln(2/a). Let's compute derivative of ln(2/a):Let me write ln(2/a) as ln 2 - ln a. Therefore, derivative is -1/a.So the second term is a * (-1/a) = -1.Therefore, the derivative of a ln(2/a) is ln(2/a) - 1.Putting it all together:k'(a) = -a/2 + ln(2/a) - 1Set k'(a) = 0:- (a)/2 + ln(2/a) - 1 = 0Rearranged:ln(2/a) = (a)/2 + 1Hmm, solving this equation for a might be tricky. Let's see. Let's denote t = a/2, then the equation becomes ln(2/(2t)) = t + 1. Simplify ln(1/t) = t + 1. So -ln t = t + 1. So ln t = -t -1. That's a transcendental equation. Maybe we can solve it numerically?Alternatively, maybe there's a specific value where this holds. Let me check a = 2. Let's test a = 2.Left side: ln(2/2) = ln(1) = 0Right side: (2)/2 + 1 = 1 + 1 = 2. Not equal. Not zero.a = 1:Left side: ln(2/1) = ln 2 ≈ 0.693Right side: 1/2 + 1 = 1.5. Not equal.a = something else. Let's try a = 1. Let's compute k'(1):k'(1) = -1/2 + ln(2/1) -1 = -0.5 + 0.693 -1 ≈ -0.5 -1 + 0.693 ≈ -1.5 + 0.693 ≈ -0.807. Negative.At a=1, derivative is negative.Let's try a=2:k'(2) = -2/2 + ln(2/2) -1 = -1 + 0 -1 = -2. Still negative.Wait, but we need to find where derivative is zero, but it's decreasing from a=1 onwards? Wait, maybe the function k(a) is decreasing for all a ≥1? But then the maximum of k(a) would be at a=1.Wait, but maybe not. Let's check a=0.5, though a must be ≥1. Wait, no. Let's think again.Wait, we are considering a ≥1. So at a=1, derivative k'(1) ≈ -0.5 + ln2 -1 ≈ -1.5 + 0.693 ≈ -0.807, which is negative. For a=2, it's -2/2 + ln(1) -1 = -1 +0 -1= -2. So derivative is always negative for a ≥1. Therefore, the function k(a) is decreasing for a ≥1. Therefore, its maximum occurs at a=1.Therefore, the maximum value of k(a) is at a=1:k(1) = 1 - (1)/4 + 1 * ln(2/1) = 1 - 0.25 + ln2 ≈ 0.75 + 0.693 ≈ 1.443. Wait, but let's compute exactly:k(1) = (4 -1)/4 -1*ln1 +1*ln2 = (3)/4 -0 + ln2 = 3/4 + ln2 ≈ 0.75 + 0.693 ≈ 1.443. Which is approximately 1.443. Hmm. But 3/4 + ln2 is exactly 3/4 + ln2. So the maximum value of k(a) for a ≥1 is 3/4 + ln2, achieved at a=1. Since the function k(a) is decreasing for a ≥1, the maximum is at a=1, and then it decreases. Therefore, the upper bound for b is 3/4 + ln2. Therefore, the range of b is all real numbers less than 3/4 + ln2. Therefore, the answer is D: (-∞, 3/4 + ln2).But let me confirm again.Wait, the problem says "determine the range of the real number b" such that "there exists a real number a (a ≥1)" making the graphs have no common points. So for each b in the range, there exists some a ≥1 such that h(x) = f(x) - g(x) is always positive (or always negative, but since h(x) tends to infinity, it's positive). Therefore, the maximum upper bound for b is 3/4 + ln2, and since a can be chosen as 1 to make the upper limit 3/4 + ln2, and for larger a, the upper limit would be smaller, so the maximum possible b is 3/4 + ln2. Hence, b must be less than this value. Therefore, the correct answer is D: (-∞, 3/4 + ln2).But wait, the answer choices are:A: [1,+∞)B: [1, 3/4 + ln2 )C: [3/4 + ln2, +∞)D: (-∞, 3/4 + ln2 )Therefore, the answer is D. But let me check if there's any possible lower bound on b. Wait, the problem states that a ≥1. So when a increases, the upper bound for b (which is k(a) = (4 -a²)/4 -a ln a + a ln2) decreases. So the maximum upper bound is at a=1, which is 3/4 + ln2. Therefore, for b to satisfy the condition, it must be less than 3/4 + ln2. There is no lower bound given in the problem. However, maybe when a approaches infinity, the upper bound for b tends to negative infinity. Therefore, for any b < 3/4 + ln2, there exists a sufficiently large a such that b is less than k(a). Wait, but when a increases, k(a) = (4 -a²)/4 -a ln a + a ln2. Let's see the behavior as a increases:The term (4 -a²)/4 tends to -a²/4, which goes to -infty.The term -a ln a tends to -infty.The term +a ln2 tends to +infty.So overall, k(a) tends to -infty as a → infinity because the -a²/4 and -a ln a dominate over +a ln2.Therefore, as a increases, the upper bound for b decreases to -infty. Therefore, for any real number b, you can choose a large enough a such that k(a) is less than b. Wait, but that's the opposite. Wait, if k(a) tends to -infty as a →infty, then for any b, no matter how negative, eventually k(a) will be less than b. But in our problem, we need to have b < k(a) for some a ≥1. Therefore, if we can find a a ≥1 such that b < k(a), then such a b is allowed. Since k(a) can be made as large as 3/4 + ln2 (at a=1) and as small as -infty (as a increases), the range of b is all real numbers less than 3/4 + ln2. Because for any b < 3/4 + ln2, we can set a=1, and then if b < k(1) = 3/4 + ln2, then it's allowed. If b is greater or equal to k(1), then even at a=1, we can't have b < k(a), and for higher a, k(a) is even smaller, so no a would work. Therefore, the range of b is all real numbers less than 3/4 + ln2. Therefore, answer D.But let me check if when a=1, the upper limit is 3/4 + ln2, which is approximately 0.75 + 0.693 = 1.443. So the upper limit is around 1.443, so the answer is D: (-infty, 3/4 + ln2). That matches.Wait, but the options given are:A: [1,+∞)B: [1, 3/4 + ln2 )C: [3/4 + ln2, +∞)D: (-∞, 3/4 + ln2 )So D is the correct answer. Therefore, the answer is D.Final Answerboxed{D}</think>Given the functions ( f(x) = x^2 - ax ) and ( g(x) = b + a ln(x - 1) ), we need to determine the range of the real number ( b ) such that there exists a real number ( a geq 1 ) where the graphs of ( y = f(x) ) and ( y = g(x) ) have no common points.1. Equation Setup: The equation ( f(x) = g(x) ) must have no real solutions for ( x > 1 ). This translates to the equation ( x^2 - ax - a ln(x - 1) - b = 0 ) having no real solutions.2. Critical Points Analysis: We analyze the function ( h(x) = x^2 - ax - a ln(x - 1) - b ). To ensure ( h(x) neq 0 ) for all ( x > 1 ), we find the critical points of ( h(x) ).3. Derivative and Critical Points: The derivative ( h'(x) = 2x - a - frac{a}{x - 1} ). Setting ( h'(x) = 0 ) gives the critical point ( x = frac{2 + a}{2} ), which is valid for ( x > 1 ).4. Minimum Value Calculation: Evaluating ( h(x) ) at ( x = frac{2 + a}{2} ): [ hleft(frac{2 + a}{2}right) = frac{4 - a^2}{4} - a ln a + a ln 2 - b ] For ( h(x) > 0 ), we need: [ b < frac{4 - a^2}{4} - a ln a + a ln 2 ]5. Maximizing the Expression: The expression ( frac{4 - a^2}{4} - a ln a + a ln 2 ) is maximized at ( a = 1 ): [ left. frac{4 - a^2}{4} - a ln a + a ln 2 right|_{a=1} = frac{3}{4} + ln 2 ]6. Conclusion: The maximum value of the expression is ( frac{3}{4} + ln 2 ). Therefore, the range of ( b ) is all real numbers less than ( frac{3}{4} + ln 2 ).The final answer is (boxed{D}).,<think>Okay, let's see. I need to find the range of the real number ( b ) such that the graphs of ( y = f(x) ) and ( y = g(x) ) have no common points, given that ( a geq 1 ). The functions are ( f(x) = x^2 - a x ) and ( g(x) = b + a ln(x - 1) ). The answer choices are given, so maybe after solving it, I can match my result to the options.First, the problem states that there's a real number ( a ) (with ( a geq 1 )) such that the two graphs don't intersect. So, for each ( a geq 1 ), we might have a certain range of ( b ) where there's no intersection, but since the question is about the existence of such an ( a ), I need to find the values of ( b ) for which there exists at least one ( a geq 1 ) making the two functions not intersect.Alternatively, maybe it's phrased as: For there to exist an ( a geq 1 ) such that ( f(x) ) and ( g(x) ) don't intersect, what must ( b ) be? So the problem is asking for the range of ( b ) where there exists some ( a geq 1 ) making the two graphs not intersect. Therefore, the complement would be: For all ( a geq 1 ), the graphs intersect, which we need to exclude.To approach this, I should set ( f(x) = g(x) ) and analyze when this equation has no real solutions. So, set ( x^2 - a x = b + a ln(x - 1) ). Let's rearrange this equation to ( x^2 - a x - a ln(x - 1) - b = 0 ). Let's define a function ( h(x) = x^2 - a x - a ln(x - 1) - b ). We need ( h(x) = 0 ) to have no solutions. Therefore, we need ( h(x) > 0 ) for all ( x ) in the domain or ( h(x) < 0 ) for all ( x ) in the domain.But first, let's note the domain of ( g(x) ). Since ( g(x) = b + a ln(x - 1) ), the argument of the logarithm must be positive, so ( x - 1 > 0 implies x > 1 ). Therefore, both ( f(x) ) and ( g(x) ) must be considered for ( x > 1 ).Therefore, we need to analyze ( h(x) = x^2 - a x - a ln(x - 1) - b ) for ( x > 1 ). We need to ensure that ( h(x) ) does not cross zero for any ( x > 1 ). To guarantee that, the minimum value of ( h(x) ) must be greater than zero or the maximum value must be less than zero. But since ( x^2 ) dominates as ( x ) approaches infinity, ( h(x) ) tends to infinity. Therefore, ( h(x) ) cannot be always negative. Hence, the only possibility is that ( h(x) ) is always positive. So, we need the minimum value of ( h(x) ) on ( x > 1 ) to be greater than zero.Therefore, we need to find the minimum of ( h(x) ) for ( x > 1 ), set it to be greater than zero, and then find the corresponding ( b ).To find the minimum, compute the derivative of ( h(x) ):( h'(x) = 2x - a - frac{a}{x - 1} ).Set ( h'(x) = 0 ):( 2x - a - frac{a}{x - 1} = 0 ).Let me solve this equation for ( x ). Let's denote ( t = x - 1 ), so ( t > 0 ), then ( x = t + 1 ). Substitute into the equation:( 2(t + 1) - a - frac{a}{t} = 0 )Simplify:( 2t + 2 - a - frac{a}{t} = 0 )Multiply both sides by ( t ) to eliminate the denominator:( 2t^2 + 2t - a t - a = 0 )Rearrange:( 2t^2 + (2 - a)t - a = 0 )This is a quadratic equation in ( t ). Let's write it as:( 2t^2 + (2 - a) t - a = 0 )The solutions are:( t = frac{ - (2 - a) pm sqrt{(2 - a)^2 + 8a} }{4} )Simplify the discriminant:( (2 - a)^2 + 8a = 4 - 4a + a^2 + 8a = a^2 + 4a + 4 = (a + 2)^2 )Therefore,( t = frac{ - (2 - a) pm (a + 2) }{4} )Compute both roots:First root:( t = frac{ - (2 - a) + (a + 2) }{4} = frac{ -2 + a + a + 2 }{4} = frac{2a}{4} = frac{a}{2} )Second root:( t = frac{ - (2 - a) - (a + 2) }{4} = frac{ -2 + a - a - 2 }{4} = frac{ -4 }{4 } = -1 )But ( t = x - 1 > 0 ), so the second root ( t = -1 ) is invalid. Therefore, the critical point is at ( t = a/2 ), so ( x = t + 1 = (a/2) + 1 ).Therefore, the minimum of ( h(x) ) occurs at ( x = 1 + a/2 ).Now, substitute ( x = 1 + a/2 ) into ( h(x) ) to find the minimum value.First, compute ( x = 1 + a/2 ).Compute each term of ( h(x) ):1. ( x^2 = left(1 + frac{a}{2}right)^2 = 1 + a + frac{a^2}{4} )2. ( -a x = -a left(1 + frac{a}{2}right) = -a - frac{a^2}{2} )3. ( -a ln(x - 1) = -a lnleft(frac{a}{2}right) )4. ( -b )Therefore, summing all terms:( h(x) = left(1 + a + frac{a^2}{4}right) + left(-a - frac{a^2}{2}right) + left(-a lnleft(frac{a}{2}right)right) - b )Simplify step by step:First, expand the terms:1. ( 1 + a + frac{a^2}{4} )2. ( -a - frac{a^2}{2} )3. ( -a lnleft(frac{a}{2}right) )4. ( -b )Combine like terms:For constants: 1For ( a ) terms: ( a - a = 0 )For ( a^2 ) terms: ( frac{a^2}{4} - frac{a^2}{2} = -frac{a^2}{4} )So:( 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) - b )Therefore, the minimum value of ( h(x) ) is ( 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) - b ).We need this minimum to be greater than zero:( 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) - b > 0 )Rearranged for ( b ):( b < 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) )So, in order for there to exist an ( a geq 1 ) such that ( h(x) > 0 ) for all ( x > 1 ), the value of ( b ) must be less than ( 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) ) for some ( a geq 1 ). Therefore, the range of ( b ) is all real numbers less than the maximum of ( 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) ) over ( a geq 1 ).Hence, we need to find the maximum value of the function ( k(a) = 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) ) for ( a geq 1 ). Then, the range of ( b ) will be ( (-infty, max k(a)) ).So, let's analyze ( k(a) ).First, rewrite ( k(a) ):( k(a) = 1 - frac{a^2}{4} - a lnleft( frac{a}{2} right) )Simplify the logarithm term:( lnleft( frac{a}{2} right) = ln a - ln 2 )Therefore:( k(a) = 1 - frac{a^2}{4} - a (ln a - ln 2) = 1 - frac{a^2}{4} - a ln a + a ln 2 )So:( k(a) = 1 + a ln 2 - frac{a^2}{4} - a ln a )Now, to find the maximum of ( k(a) ) for ( a geq 1 ), we can take the derivative of ( k(a) ) with respect to ( a ), set it equal to zero, and solve for ( a ).Compute ( k'(a) ):( k'(a) = ln 2 - frac{2a}{4} - ln a - a cdot frac{1}{a} )Simplify each term:1. The derivative of ( a ln 2 ) is ( ln 2 ).2. The derivative of ( -frac{a^2}{4} ) is ( -frac{2a}{4} = -frac{a}{2} ).3. The derivative of ( -a ln a ) is ( - ln a - a cdot frac{1}{a} = - ln a - 1 ).Therefore, combining these:( k'(a) = ln 2 - frac{a}{2} - ln a - 1 )Set ( k'(a) = 0 ):( ln 2 - frac{a}{2} - ln a - 1 = 0 )Let's rearrange:( - frac{a}{2} - ln a + (ln 2 - 1) = 0 )Multiply both sides by -1:( frac{a}{2} + ln a - (ln 2 - 1) = 0 )Simplify ( - (ln 2 - 1) = 1 - ln 2 ):( frac{a}{2} + ln a + 1 - ln 2 = 0 )But this seems complicated. Let me check my steps again.Wait, perhaps I made a mistake in the derivative. Let me verify:Original ( k(a) = 1 + a ln 2 - frac{a^2}{4} - a ln a )Derivative term by term:- The derivative of 1 is 0.- The derivative of ( a ln 2 ) is ( ln 2 ).- The derivative of ( -frac{a^2}{4} ) is ( -frac{2a}{4} = -frac{a}{2} ).- The derivative of ( -a ln a ) is ( - [ ln a + a cdot frac{1}{a} ] = - ln a - 1 ).Yes, so total derivative is ( ln 2 - frac{a}{2} - ln a - 1 ). So that's correct.Setting derivative to zero:( ln 2 - frac{a}{2} - ln a - 1 = 0 )Let me write this as:( - frac{a}{2} - ln a = 1 - ln 2 )Multiply both sides by -1:( frac{a}{2} + ln a = ln 2 - 1 )Wait, ( ln 2 - 1 ) is approximately ( 0.6931 - 1 = -0.3069 ). But the left-hand side ( frac{a}{2} + ln a ) is for ( a geq 1 ). Let's check at ( a = 1 ):Left-hand side: ( frac{1}{2} + ln 1 = 0.5 + 0 = 0.5 ). Right-hand side: ( ln 2 - 1 approx -0.3069 ). So ( 0.5 neq -0.3069 ). Hmm. So the equation ( frac{a}{2} + ln a = ln 2 - 1 ) may not have a solution because the left-hand side is increasing for ( a geq 1 ). Wait, let's check if the left-hand side can ever be negative.At ( a = 1 ): LHS = 0.5. As ( a ) increases, ( frac{a}{2} ) increases and ( ln a ) increases. So LHS is increasing. Therefore, the LHS is always greater than or equal to 0.5 for ( a geq 1 ), but the RHS is negative. Therefore, there's no solution to ( frac{a}{2} + ln a = ln 2 - 1 ). Therefore, the equation ( k'(a) = 0 ) has no solution for ( a geq 1 ).Therefore, the function ( k(a) ) is decreasing for all ( a geq 1 ), because the derivative ( k'(a) = ln 2 - frac{a}{2} - ln a - 1 ) is always negative for ( a geq 1 ). Let me check the derivative at ( a = 1 ):( k'(1) = ln 2 - frac{1}{2} - ln 1 - 1 = ln 2 - 0.5 - 0 - 1 = ln 2 - 1.5 approx 0.6931 - 1.5 approx -0.8069 ), which is negative.Since the derivative is negative at ( a = 1 ) and becomes more negative as ( a ) increases (because the term ( - frac{a}{2} ) becomes more negative), ( k(a) ) is decreasing for all ( a geq 1 ). Therefore, the maximum of ( k(a) ) occurs at ( a = 1 ).Therefore, the maximum value of ( k(a) ) for ( a geq 1 ) is ( k(1) ).Compute ( k(1) ):( k(1) = 1 - frac{1^2}{4} - 1 cdot lnleft( frac{1}{2} right) )Simplify:( 1 - frac{1}{4} - lnleft( frac{1}{2} right) )( = frac{3}{4} - (-ln 2) )( = frac{3}{4} + ln 2 )Therefore, ( k(a) ) attains its maximum value of ( frac{3}{4} + ln 2 ) at ( a = 1 ), and decreases as ( a ) increases beyond 1. Therefore, the maximum value of ( k(a) ) over ( a geq 1 ) is ( frac{3}{4} + ln 2 ), and for ( b ) to satisfy ( b < k(a) ) for some ( a geq 1 ), the range of ( b ) is all real numbers less than ( frac{3}{4} + ln 2 ). Therefore, the answer is ( (-infty, frac{3}{4} + ln 2) ), which is option D.But wait, let me check again. The problem states "there exists a real number ( a (a geq 1) ) such that the graph of ( y = f(x) ) has no common points with the graph of ( y = g(x) ). Determine the range of the real number ( b )." So, if ( b < frac{3}{4} + ln 2 ), then there exists an ( a geq 1 ) (specifically, when ( a = 1 ), and maybe other ( a )) such that the two graphs don't intersect. Conversely, if ( b geq frac{3}{4} + ln 2 ), then for all ( a geq 1 ), the graphs will intersect, because even the maximum possible ( k(a) ) is ( frac{3}{4} + ln 2 ), so ( b ) being equal or larger would mean ( h(x) ) would have a minimum less than or equal to zero, leading to intersection. Therefore, the correct range for ( b ) is ( (-infty, frac{3}{4} + ln 2) ), which is option D.But let me verify with an example. Take ( a = 1 ), then the critical point is at ( x = 1 + 1/2 = 1.5 ). Compute the minimum value ( h(1.5) = 1 - 1/4 - 1 cdot ln(1/2) = 3/4 + ln 2 ). So, if ( b < 3/4 + ln 2 ), then when ( a = 1 ), the minimum of ( h(x) ) is positive, so no intersection. If ( b geq 3/4 + ln 2 ), then even for ( a = 1 ), the minimum is less than or equal to zero, so there would be an intersection. However, for ( a > 1 ), since ( k(a) ) is decreasing, the maximum value of ( k(a) ) is smaller. Therefore, even if ( b ) is less than some ( k(a) ) for a larger ( a ), but since the maximum occurs at ( a = 1 ), the key threshold is ( 3/4 + ln 2 ). Therefore, the answer is D.Final Answerboxed{D}</think>Given the functions ( f(x) = x^2 - ax ) and ( g(x) = b + a ln(x - 1) ), we need to determine the range of the real number ( b ) such that there exists a real number ( a geq 1 ) where the graphs of ( y = f(x) ) and ( y = g(x) ) have no common points.To solve this, we set ( f(x) = g(x) ) and analyze when this equation has no real solutions. We define the function ( h(x) = x^2 - ax - a ln(x - 1) - b ) and need ( h(x) > 0 ) for all ( x > 1 ). First, we find the critical points by taking the derivative of ( h(x) ):[h'(x) = 2x - a - frac{a}{x - 1}]Setting ( h'(x) = 0 ) and solving, we find the critical point at ( x = 1 + frac{a}{2} ).Substituting ( x = 1 + frac{a}{2} ) into ( h(x) ), we compute the minimum value of ( h(x) ):[hleft(1 + frac{a}{2}right) = 1 - frac{a^2}{4} - a lnleft(frac{a}{2}right) - b]We need this minimum value to be positive:[1 - frac{a^2}{4} - a lnleft(frac{a}{2}right) - b > 0 implies b < 1 - frac{a^2}{4} - a lnleft(frac{a}{2}right)]To find the maximum value of the right-hand side for ( a geq 1 ), we analyze the function ( k(a) = 1 - frac{a^2}{4} - a lnleft(frac{a}{2}right) ). Taking the derivative and solving, we find that ( k(a) ) is decreasing for ( a geq 1 ). The maximum value occurs at ( a = 1 ):[k(1) = 1 - frac{1}{4} - lnleft(frac{1}{2}right) = frac{3}{4} + ln 2]Thus, the range of ( b ) is all real numbers less than ( frac{3}{4} + ln 2 ). The answer is (boxed{D}).]

🔑:If the graphs of y=f(x) and y=g(x) have no common points, then it is equivalent to f(x)-g(x) > 0 or f(x)-g(x) 0 or x^{2}-ax-b-aln (x-1) 0.This means that the function h(x) attains its minimum value (which is also the global minimum) at x= frac {a+2}{2}, and this minimum value is h( frac {a+2}{2})=- frac {a^{2}}{4}+1-aln frac {a}{2}.Let G(a)=h( frac {a+2}{2})=- frac {a^{2}}{4}+1-aln frac {a}{2}.Then, G(a) is a decreasing function on [1,+infty).Therefore, the maximum value of G(a) is G(1)= frac {3}{4}+ln 2.Hence, the minimum value of h(x), h( frac {a+2}{2})leqslant frac {3}{4}+ln 2.If x^{2}-ax-aln (x-1) > b, then b < frac {3}{4}+ln 2.If x^{2}-ax-aln (x-1) < b is always true, then this is not feasible.In conclusion, b < frac {3}{4}+ln 2.Therefore, the correct answer is: boxed{D}.

🧠:[<think>Okay, so I have this problem here: If (frac{b}{a} = frac{c}{b} = frac{2}{c}), then I need to find the value of the expression (b^{2} - c^{2} - a c + b c - 2 a + 2 b). Hmm, let me try to figure this out step by step.First, the problem gives me three ratios that are all equal: (frac{b}{a}), (frac{c}{b}), and (frac{2}{c}). That means all these fractions are equal to the same constant. Let me call that constant (k). So, I can write:[frac{b}{a} = k,quad frac{c}{b} = k,quad text{and}quad frac{2}{c} = k]From these equations, I can express (b), (c), and then relate them back to (a). Let me start with the first ratio: (frac{b}{a} = k), which implies that (b = a k).Next, the second ratio: (frac{c}{b} = k), so (c = b k). But since I already know that (b = a k), substituting that in gives (c = a k cdot k = a k^2).Then, the third ratio: (frac{2}{c} = k), which means (c = frac{2}{k}). But wait, from the previous step, I have (c = a k^2). So, setting these equal:[a k^2 = frac{2}{k}]Solving for (a), multiply both sides by (k):[a k^3 = 2 implies a = frac{2}{k^3}]Okay, so now I have expressions for (a), (b), and (c) in terms of (k):- (a = frac{2}{k^3})- (b = a k = frac{2}{k^3} cdot k = frac{2}{k^2})- (c = a k^2 = frac{2}{k^3} cdot k^2 = frac{2}{k})So, now all variables are expressed in terms of (k). Maybe I can find the value of (k) first?Let me check if there's another equation I can use. Wait, all three ratios are equal to (k), so maybe there's a way to find (k) directly.Starting from the first ratio: (b = a k), second ratio: (c = b k), third ratio: (2 = c k). Let's substitute each into the next:From the first to the second: (c = b k = (a k) k = a k^2).From the third ratio: (2 = c k = (a k^2) k = a k^3). But we already found that (a = frac{2}{k^3}), so substituting back into (2 = a k^3):(2 = frac{2}{k^3} cdot k^3 = 2). Hmm, so that's just an identity, which means there's no new information here, and (k) can be any value? Wait, that can't be right. Maybe I missed something here.Wait, but all three ratios are equal, so maybe there's a way to relate them without introducing (k). Let me see:Given (frac{b}{a} = frac{c}{b} = frac{2}{c}). Let's set all equal to some constant (k), but perhaps instead of using (k), we can set up equations from the equalities.First, (frac{b}{a} = frac{c}{b}). Cross-multiplying gives (b^2 = a c).Second, (frac{c}{b} = frac{2}{c}). Cross-multiplying gives (c^2 = 2 b).So now, from the second equation, (c^2 = 2 b). Let's keep that in mind.From the first equation, (b^2 = a c). So, if we can express (a) in terms of (b) or (c), that might help.From the first ratio, (frac{b}{a} = frac{2}{c}), since all three ratios are equal. Wait, actually, all three ratios are equal, so (frac{b}{a} = frac{2}{c}) as well. So that gives another equation: (frac{b}{a} = frac{2}{c}), so cross-multiplying gives (b c = 2 a).So now, we have three equations:1. (b^2 = a c)2. (c^2 = 2 b)3. (b c = 2 a)Let me see if we can solve these equations for (a), (b), (c).From equation 3: (2 a = b c implies a = frac{b c}{2}).Substitute this into equation 1: (b^2 = a c = frac{b c}{2} cdot c = frac{b c^2}{2}).Therefore:[b^2 = frac{b c^2}{2} implies b = frac{c^2}{2}]But from equation 2, we have (c^2 = 2 b). Substitute that into the above equation:[b = frac{2 b}{2} = b]Hmm, which simplifies to (b = b), which is always true. So again, this seems like we need another way to relate these variables.Wait, perhaps substituting equation 2 into the expression for (a). Let's try:From equation 2: (c^2 = 2 b implies b = frac{c^2}{2}).Then, from equation 3: (a = frac{b c}{2} = frac{frac{c^2}{2} cdot c}{2} = frac{c^3}{4}).So now, we have (a = frac{c^3}{4}) and (b = frac{c^2}{2}).So all variables can be expressed in terms of (c). Maybe we can find (c)?But how? Let's check if there's another relation. Wait, from the ratios:(frac{b}{a} = frac{2}{c}). Let's substitute (a = frac{c^3}{4}) and (b = frac{c^2}{2}) into this ratio.Left-hand side: (frac{b}{a} = frac{frac{c^2}{2}}{frac{c^3}{4}} = frac{c^2}{2} cdot frac{4}{c^3} = frac{4}{2 c} = frac{2}{c}).Right-hand side: (frac{2}{c}).So that checks out. Therefore, the expressions for (a) and (b) in terms of (c) are consistent. But since there are no more equations, it seems like (c) can be any value? But that contradicts the problem statement, which asks for a specific value of the expression. Therefore, perhaps the expression (b^{2} - c^{2} - a c + b c - 2 a + 2 b) can be simplified in terms of (c) (or (k)) and when simplified, the variables cancel out, leaving a constant.Alternatively, since all variables are expressed in terms of (k), maybe substituting back into the expression will result in a numerical value. Let's try that approach.Earlier, I had expressed all variables in terms of (k):- (a = frac{2}{k^3})- (b = frac{2}{k^2})- (c = frac{2}{k})So, substitute these into the expression (b^{2} - c^{2} - a c + b c - 2 a + 2 b):First, compute each term:1. (b^2 = left(frac{2}{k^2}right)^2 = frac{4}{k^4})2. (c^2 = left(frac{2}{k}right)^2 = frac{4}{k^2})3. (a c = frac{2}{k^3} cdot frac{2}{k} = frac{4}{k^4})4. (b c = frac{2}{k^2} cdot frac{2}{k} = frac{4}{k^3})5. (2 a = 2 cdot frac{2}{k^3} = frac{4}{k^3})6. (2 b = 2 cdot frac{2}{k^2} = frac{4}{k^2})Now plug these into the expression:[b^{2} - c^{2} - a c + b c - 2 a + 2 b = frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2}]Let me simplify term by term:1. ( frac{4}{k^4} - frac{4}{k^4} = 0 )2. ( - frac{4}{k^2} + frac{4}{k^2} = 0 )3. ( frac{4}{k^3} - frac{4}{k^3} = 0 )So all terms cancel out, resulting in 0.Wait, that's unexpected. So the entire expression simplifies to 0 regardless of the value of (k). Therefore, the answer should be 0?But let me verify this again step by step to ensure I didn't make a mistake in substitution or arithmetic.Given:- (b^{2} = frac{4}{k^4})- (c^{2} = frac{4}{k^2})- (a c = frac{4}{k^4})- (b c = frac{4}{k^3})- (2 a = frac{4}{k^3})- (2 b = frac{4}{k^2})So plugging into the expression:(b^2 - c^2 - a c + b c - 2a + 2b)Translates to:(frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2})Group the terms:- (frac{4}{k^4} - frac{4}{k^4} = 0)- (- frac{4}{k^2} + frac{4}{k^2} = 0)- (frac{4}{k^3} - frac{4}{k^3} = 0)Thus, all terms indeed cancel out, leaving 0.Therefore, regardless of the value of (k), the expression equals 0. Therefore, the answer is 0.Wait, but let me check with actual numbers. Maybe pick a specific value for (k) and compute all variables and the expression to confirm.Suppose (k = 1). Then:- (a = 2/1^3 = 2)- (b = 2/1^2 = 2)- (c = 2/1 = 2)But let's check if the ratios hold:(frac{b}{a} = 2/2 = 1), (frac{c}{b} = 2/2 = 1), (frac{2}{c} = 2/2 = 1). Okay, that works.Now compute the expression:(b^2 - c^2 - a c + b c - 2a + 2b)Which is:(2^2 - 2^2 - 2*2 + 2*2 - 2*2 + 2*2)Calculate each term:- (4 - 4 - 4 + 4 - 4 + 4)Simplify:(4 - 4) = 00 - 4 = -4-4 + 4 = 00 - 4 = -4-4 + 4 = 0Result is 0. So that works.Let me try another value for (k), say (k = 2).Then:- (a = 2/(2)^3 = 2/8 = 1/4)- (b = 2/(2)^2 = 2/4 = 1/2)- (c = 2/2 = 1)Check ratios:(frac{b}{a} = (1/2)/(1/4) = 2), (frac{c}{b} = 1/(1/2) = 2), (frac{2}{c} = 2/1 = 2). Good.Now compute the expression:(b^2 - c^2 - a c + b c - 2a + 2b)Compute each term:- (b^2 = (1/2)^2 = 1/4)- (c^2 = 1^2 = 1)- (a c = (1/4)(1) = 1/4)- (b c = (1/2)(1) = 1/2)- (2 a = 2*(1/4) = 1/2)- (2 b = 2*(1/2) = 1)Putting them together:(1/4 - 1 - 1/4 + 1/2 - 1/2 + 1)Simplify step by step:1. (1/4 - 1 = -3/4)2. (-3/4 - 1/4 = -1)3. (-1 + 1/2 = -1/2)4. (-1/2 - 1/2 = -1)5. (-1 + 1 = 0)Again, the result is 0. Hmm, so regardless of (k), the expression evaluates to 0. So the answer is indeed 0.Alternatively, maybe there's a way to simplify the original expression without substituting variables. Let me try that approach to verify.Given the expression:(b^2 - c^2 - a c + b c - 2 a + 2 b)Let me rearrange the terms:Group (b^2 - c^2) and then the rest:((b^2 - c^2) + (- a c + b c) + (- 2 a + 2 b))Factor each group:1. (b^2 - c^2 = (b - c)(b + c))2. (-a c + b c = c(-a + b) = c(b - a))3. (-2 a + 2 b = 2(b - a))So the expression becomes:((b - c)(b + c) + c(b - a) + 2(b - a))Notice that (b - a) can be factored from the last two terms:((b - c)(b + c) + (b - a)(c + 2))Now, let's recall from the given ratios:From (frac{b}{a} = frac{c}{b}), cross-multiplied gives (b^2 = a c). So (a = frac{b^2}{c}).Let me substitute (a = frac{b^2}{c}) into the expression:First, compute (b - a = b - frac{b^2}{c} = frac{b c - b^2}{c} = frac{b(c - b)}{c})Then, substitute into the expression:((b - c)(b + c) + left(frac{b(c - b)}{c}right)(c + 2))Note that (b - c = - (c - b)), so:(- (c - b)(b + c) + frac{b(c - b)(c + 2)}{c})Factor out ( (c - b) ):( (c - b) left[ - (b + c) + frac{b(c + 2)}{c} right] )Simplify the expression inside the brackets:First, distribute the terms:( -b - c + frac{b(c + 2)}{c} )Break down the fraction:( -b - c + frac{b c + 2 b}{c} = -b - c + b + frac{2 b}{c} )Simplify:- ( -b + b = 0 )- ( -c + frac{2 b}{c} )So the entire expression becomes:( (c - b) left( -c + frac{2 b}{c} right) )Now, let's simplify this:Multiply ( -c ) and ( frac{2 b}{c} ):( (c - b) left( -c + frac{2 b}{c} right) = (c - b) left( frac{ - c^2 + 2 b }{c} right) )Multiply numerator terms:( frac{(c - b)( - c^2 + 2 b )}{c} )Hmm, this seems complicated, but maybe we can use another relation. Remember that from the ratios, we had (c^2 = 2 b). Let's use that here.Given (c^2 = 2 b), so ( - c^2 + 2 b = -2 b + 2 b = 0 ).Therefore, the entire expression becomes:( frac{(c - b)(0)}{c} = 0 )So the original expression simplifies to 0.Therefore, regardless of the method used, substitution or algebraic manipulation, the expression evaluates to 0. Therefore, the answer is 0.Final Answerboxed{0}</think>Given the problem where (frac{b}{a} = frac{c}{b} = frac{2}{c}), we need to find the value of the expression (b^2 - c^2 - ac + bc - 2a + 2b).First, we introduce a constant (k) such that:[frac{b}{a} = k, quad frac{c}{b} = k, quad frac{2}{c} = k]From these, we express (a), (b), and (c) in terms of (k):[b = ak, quad c = bk, quad c = frac{2}{k}]Substituting (b = ak) into (c = bk), we get:[c = ak^2]Equating this with (c = frac{2}{k}), we find:[ak^2 = frac{2}{k} implies a = frac{2}{k^3}]Thus, substituting back, we have:[a = frac{2}{k^3}, quad b = frac{2}{k^2}, quad c = frac{2}{k}]Next, we substitute these values into the expression (b^2 - c^2 - ac + bc - 2a + 2b):1. (b^2 = left(frac{2}{k^2}right)^2 = frac{4}{k^4})2. (c^2 = left(frac{2}{k}right)^2 = frac{4}{k^2})3. (ac = frac{2}{k^3} cdot frac{2}{k} = frac{4}{k^4})4. (bc = frac{2}{k^2} cdot frac{2}{k} = frac{4}{k^3})5. (2a = 2 cdot frac{2}{k^3} = frac{4}{k^3})6. (2b = 2 cdot frac{2}{k^2} = frac{4}{k^2})Substituting these into the expression:[frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2}]Simplifying term by term:[left(frac{4}{k^4} - frac{4}{k^4}right) + left(-frac{4}{k^2} + frac{4}{k^2}right) + left(frac{4}{k^3} - frac{4}{k^3}right) = 0]Thus, the value of the expression is:[boxed{0}],<think>Okay, so I have this problem here: If (frac{b}{a} = frac{c}{b} = frac{2}{c}), then I need to find the value of the expression (b^{2} - c^{2} - ac + bc - 2a + 2b). Hmm, let's try to figure this out step by step.First, the problem gives me a ratio equality: (frac{b}{a} = frac{c}{b} = frac{2}{c}). That means all these fractions are equal to each other. Let's call that common ratio (k). So, I can write:[frac{b}{a} = k,quad frac{c}{b} = k,quad frac{2}{c} = k]From these equations, I can express each variable in terms of (k). Let's start from the first one: (frac{b}{a} = k) implies (b = a k).Then, the second equation (frac{c}{b} = k) gives (c = b k). But since (b = a k), substitute that in: (c = (a k) cdot k = a k^2).The third equation is (frac{2}{c} = k), so solving for (c) gives (c = frac{2}{k}).Wait, so now I have two expressions for (c): one is (c = a k^2) and the other is (c = frac{2}{k}). Let me set them equal to each other:[a k^2 = frac{2}{k}]Solving for (a), multiply both sides by (k):[a k^3 = 2 implies a = frac{2}{k^3}]Alright, so (a = frac{2}{k^3}). Then, since (b = a k), substitute (a) into that:[b = frac{2}{k^3} cdot k = frac{2}{k^2}]And we already have (c = frac{2}{k}).So now, all variables (a), (b), (c) are expressed in terms of (k). Maybe I can substitute these into the expression (b^{2} - c^{2} - ac + bc - 2a + 2b) and simplify?Let's write down each term:First, (b^2 = left(frac{2}{k^2}right)^2 = frac{4}{k^4}).Next, (c^2 = left(frac{2}{k}right)^2 = frac{4}{k^2}).Then, (ac = frac{2}{k^3} cdot frac{2}{k} = frac{4}{k^4}).Then, (bc = frac{2}{k^2} cdot frac{2}{k} = frac{4}{k^3}).Next, (2a = 2 cdot frac{2}{k^3} = frac{4}{k^3}).Finally, (2b = 2 cdot frac{2}{k^2} = frac{4}{k^2}).Now plug these into the expression:[b^{2} - c^{2} - ac + bc - 2a + 2b = frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2}]Let's combine like terms:- The first term is (frac{4}{k^4}), and the third term is (-frac{4}{k^4}), so they cancel out.- The second term is (-frac{4}{k^2}) and the last term is (+frac{4}{k^2}), so those also cancel each other.- The fourth term is (+frac{4}{k^3}) and the fifth term is (-frac{4}{k^3}), which also cancel.So after simplifying, everything cancels out, leaving us with 0.Wait, that's surprising. Let me check my substitutions again to see if I made any mistakes.Starting from the beginning:- (a = frac{2}{k^3})- (b = frac{2}{k^2})- (c = frac{2}{k})Compute (b^2):(left(frac{2}{k^2}right)^2 = frac{4}{k^4}) – correct.Compute (c^2):(left(frac{2}{k}right)^2 = frac{4}{k^2}) – correct.Compute (ac):(frac{2}{k^3} times frac{2}{k} = frac{4}{k^4}) – correct.Compute (bc):(frac{2}{k^2} times frac{2}{k} = frac{4}{k^3}) – correct.Compute (2a):(2 times frac{2}{k^3} = frac{4}{k^3}) – correct.Compute (2b):(2 times frac{2}{k^2} = frac{4}{k^2}) – correct.So substituting back into the expression:[frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2}]Now, step by step:1. (frac{4}{k^4} - frac{4}{k^4} = 0)2. (-frac{4}{k^2} + frac{4}{k^2} = 0)3. (frac{4}{k^3} - frac{4}{k^3} = 0)So all terms cancel, resulting in 0. Hmm. Is that possible? Maybe the expression simplifies to zero regardless of the value of (k). Let me check with actual numbers.Suppose we choose a value for (k). Let's find (a), (b), (c) in terms of (k) and compute the expression.Wait, but since all terms cancel out in terms of (k), the expression must always equal zero. Therefore, the answer is 0.Alternatively, maybe there's another approach. Let's see.Alternatively, since (frac{b}{a} = frac{c}{b} = frac{2}{c} = k), then the variables form a geometric progression. Let me verify.If (frac{b}{a} = frac{c}{b}), then (b^2 = a c), which is the property of a geometric progression. So, (a), (b), (c) are in geometric progression. Then, since (frac{2}{c} = k), so (c = frac{2}{k}). Then, since (b^2 = a c), we can write (a = frac{b^2}{c}).But let's see if that helps with the expression.The expression is (b^2 - c^2 - a c + b c - 2a + 2b).If (a c = b^2), then substituting (a c) with (b^2), the expression becomes:(b^2 - c^2 - b^2 + b c - 2a + 2b)Simplify:( - c^2 + b c - 2a + 2b )Hmm, maybe that's helpful. Let's see.We still need to express (a) in terms of (b) and (c). Since (a = frac{b^2}{c}), substitute that in:( -c^2 + b c - 2 cdot frac{b^2}{c} + 2b )Hmm, not sure if that's easier. Let me compute each term:- (-c^2)- (+ b c)- (- frac{2 b^2}{c})- (+ 2b)To combine these, maybe get a common denominator. Multiply terms by (c):- (-c^3 + b c^2 - 2 b^2 + 2 b c)Not sure. Alternatively, let's try substituting the values again. Wait, if all variables are expressed in terms of (k), and substitution led to 0, perhaps the answer is indeed 0.Alternatively, maybe check with specific numbers. Let me pick a value for (k) and compute.Let me choose (k = 1). Wait, but if (k = 1), then from (frac{2}{c} = 1), so (c = 2). Then (c = a k^2), so (2 = a *1^2), so (a = 2). Then (b = a k = 2*1 = 2). But then check the first ratio: (b/a = 2/2 = 1), second ratio (c/b = 2/2 = 1), third ratio (2/c = 2/2 =1). So that works.So with (a=2), (b=2), (c=2), compute the expression:(b^2 - c^2 - a c + b c - 2a + 2b)Substitute:(2^2 - 2^2 - 2*2 + 2*2 - 2*2 + 2*2)Compute:4 - 4 - 4 + 4 - 4 + 4Which is 0. So indeed, the expression is 0 here.Let me try another value for (k). Let's pick (k = 2).Then from (c = frac{2}{k}), (c = 1). Then (b = frac{2}{k^2} = frac{2}{4} = 0.5). Then (a = frac{2}{k^3} = frac{2}{8} = 0.25).Check the ratios:(b/a = 0.5 / 0.25 = 2), (c/b = 1 / 0.5 = 2), (2/c = 2 /1 = 2). Correct.Now compute the expression:(b^2 - c^2 - a c + b c - 2a + 2b)Substitute:(0.5^2 - 1^2 - 0.25*1 + 0.5*1 - 2*0.25 + 2*0.5)Compute each term:0.25 - 1 - 0.25 + 0.5 - 0.5 + 1Combine:0.25 -1 = -0.75-0.75 -0.25 = -1-1 + 0.5 = -0.5-0.5 -0.5 = -1-1 +1 = 0Again, it's zero. So this seems consistent. Therefore, regardless of the value of (k), the expression evaluates to 0.Therefore, the answer is (boxed{0}).Final Answerboxed{0}</think>Given the problem: If (frac{b}{a} = frac{c}{b} = frac{2}{c}), we need to find the value of the expression (b^{2} - c^{2} - ac + bc - 2a + 2b).First, we denote the common ratio by (k). Thus, we have:[frac{b}{a} = k, quad frac{c}{b} = k, quad frac{2}{c} = k]From these ratios, we express (a), (b), and (c) in terms of (k):- (b = a k)- (c = b k = a k^2)- (c = frac{2}{k})Equating the two expressions for (c):[a k^2 = frac{2}{k} implies a k^3 = 2 implies a = frac{2}{k^3}]Substituting (a) back, we get:- (b = frac{2}{k^3} cdot k = frac{2}{k^2})- (c = frac{2}{k})Next, we substitute these values into the expression (b^{2} - c^{2} - ac + bc - 2a + 2b):- (b^2 = left(frac{2}{k^2}right)^2 = frac{4}{k^4})- (c^2 = left(frac{2}{k}right)^2 = frac{4}{k^2})- (ac = frac{2}{k^3} cdot frac{2}{k} = frac{4}{k^4})- (bc = frac{2}{k^2} cdot frac{2}{k} = frac{4}{k^3})- (2a = 2 cdot frac{2}{k^3} = frac{4}{k^3})- (2b = 2 cdot frac{2}{k^2} = frac{4}{k^2})Substituting these into the expression:[frac{4}{k^4} - frac{4}{k^2} - frac{4}{k^4} + frac{4}{k^3} - frac{4}{k^3} + frac{4}{k^2}]Combining like terms:- (frac{4}{k^4} - frac{4}{k^4} = 0)- (-frac{4}{k^2} + frac{4}{k^2} = 0)- (frac{4}{k^3} - frac{4}{k^3} = 0)All terms cancel out, resulting in 0. Verifying with specific values of (k) (e.g., (k = 1) and (k = 2)) confirms the expression evaluates to 0.Thus, the final answer is (boxed{0}).]

🔑:Reference answer: 0

🧠:[<think>Okay, so I need to figure out which of the given equations has a constant term of zero. Let me start by recalling what a constant term is. In a quadratic equation of the form ax² + bx + c = 0, the constant term is 'c'. So, the problem is asking which equation, when rewritten in standard form (ax² + bx + c = 0), has c = 0. That means after simplifying all the equations, the one without a constant term (the term without x) is the correct answer. Let's check each option one by one.Starting with option A: x² + x = 0. Hmm, this is already in standard form. Here, a = 1, b = 1, and c = 0. So, the constant term is indeed zero. That looks like a possible candidate. But let me check the others to be sure.Option B: 2x² - x - 12 = 0. This is also in standard form. The coefficients are a = 2, b = -1, c = -12. The constant term here is -12, not zero. So B is out.Option C: 2(x² - 1) = 3(x - 1). Let me expand this equation to standard form. First, distribute the left side: 2x² - 2. On the right side, distribute the 3: 3x - 3. So the equation becomes 2x² - 2 = 3x - 3. Now, bring all terms to the left side by subtracting 3x and adding 3: 2x² - 3x + (-2 + 3) = 0. That simplifies to 2x² - 3x + 1 = 0. So here, the constant term is 1. Therefore, C is not the answer either.Option D: 2(x² + 1) = x + 4. Let me expand this one. Distribute the left side: 2x² + 2. The equation is 2x² + 2 = x + 4. Now, move all terms to the left: 2x² - x + 2 - 4 = 0. Simplify that to 2x² - x - 2 = 0. The constant term here is -2. So D is also not the correct answer.Wait, but let me double-check my steps for each option to make sure I didn't make a mistake. Starting again with A: x² + x = 0. Yes, that's straightforward. No constant term, so c = 0. So A is a possible answer. But let me check C again because sometimes when expanding, errors can occur.For option C: 2(x² - 1) = 3(x - 1). Let's do the left side: 2*x² - 2*1 = 2x² - 2. Right side: 3*x - 3*1 = 3x - 3. So equation is 2x² - 2 = 3x - 3. Subtract 3x and add 3 to both sides: 2x² - 3x + 1 = 0. Yes, that's correct. So the constant term is 1.Option D: 2(x² +1) = x +4. Left side: 2x² + 2. Right side: x +4. Subtract x and 4: 2x² - x +2 -4 = 2x² -x -2 =0. So constant term is -2. Correct.Therefore, among all options, only A has a constant term of zero. So the answer must be A. But wait, let me check the original problem statement again. It says "has a constant term of zero". So when the equation is in standard form, the constant term (c) is zero. So A: x² + x =0, which is already standard form with c=0. Therefore, the answer is A. But wait, the options given in Chinese parentheses are A, B, C, D. The user hasn't mentioned if multiple answers are possible, but the way the question is phrased is "which of the following has...", implying only one. Let me confirm once more.Yes, A's constant term is zero. B's is -12, C's is 1, D's is -2. So yes, only A. Therefore, the correct answer is option A.Final Answerboxed{A}</think>To determine which equation has a constant term of zero, we need to rewrite each equation in standard form (ax^2 + bx + c = 0) and identify the constant term (c).- Option A: (x^2 + x = 0) - Already in standard form. Here, (a = 1), (b = 1), and (c = 0). The constant term is 0.- Option B: (2x^2 - x - 12 = 0) - Already in standard form. Here, (a = 2), (b = -1), and (c = -12). The constant term is -12.- Option C: (2(x^2 - 1) = 3(x - 1)) - Expand and simplify: - Left side: (2x^2 - 2) - Right side: (3x - 3) - Equation becomes (2x^2 - 2 = 3x - 3) - Rearrange to standard form: (2x^2 - 3x + 1 = 0). The constant term is 1.- Option D: (2(x^2 + 1) = x + 4) - Expand and simplify: - Left side: (2x^2 + 2) - Right side: (x + 4) - Equation becomes (2x^2 + 2 = x + 4) - Rearrange to standard form: (2x^2 - x - 2 = 0). The constant term is -2.Among the options, only A has a constant term of zero.(boxed{A}),<think>Okay, let's see. I need to figure out which of the given quadratic equations has a constant term of zero. The options are A, B, C, D. First, I should recall what a constant term is. In a quadratic equation of the form ax² + bx + c = 0, the constant term is the term without any variable, which is 'c'. So, if the constant term is zero, that means 'c' should be zero in the standard form. Therefore, the equation should look like ax² + bx = 0. Let me check each option one by one.Starting with option A: x² + x = 0. Let's write this in standard form. It's already there: 1x² + 1x + 0 = 0. So here, a=1, b=1, c=0. That means the constant term is zero. So A seems to have a constant term of zero. But let me check the others to be sure.Option B: 2x² - x -12 =0. In standard form, that's 2x² - x -12 =0. So here, a=2, b=-1, c=-12. The constant term is -12, which isn't zero. So B is out.Option C: 2(x² -1) = 3(x -1). Let me expand this. Left side: 2x² - 2. Right side: 3x - 3. Bring all terms to the left side: 2x² - 2 -3x +3 = 0. Simplify: 2x² -3x +1 =0. So here, a=2, b=-3, c=1. Constant term is 1. Not zero. So C is not the answer.Option D: 2(x² +1) = x +4. Let's expand left side: 2x² +2. Right side: x +4. Bring all terms to the left: 2x² +2 -x -4 =0. Simplify: 2x² -x -2 =0. So here, a=2, b=-1, c=-2. The constant term is -2. Not zero. So D is out.Therefore, the only equation with a constant term of zero is option A. But wait, let me double-check my expansion for options C and D to be sure I didn't make a mistake.For option C: 2(x² -1) is 2x² -2, and 3(x -1) is 3x -3. Subtracting those gives 2x² -2 -3x +3. That simplifies to 2x² -3x +1. Yep, that's correct. So constant term 1.For option D: 2(x² +1) is 2x² +2, and moving x +4 to the left gives -x -4. So combined, 2x² +2 -x -4 = 2x² -x -2. So the constant term is -2. Correct.So, yes, only option A has a constant term of zero. The answer should be A.Final Answerboxed{A}</think>To determine which equation has a constant term of zero, we need to convert each equation to the standard quadratic form (ax^2 + bx + c = 0) and identify the constant term (c).- Option A: (x^2 + x = 0) - Standard form: (x^2 + x + 0 = 0) - Here, (a = 1), (b = 1), and (c = 0). - Constant term (c) is 0.- Option B: (2x^2 - x - 12 = 0) - Standard form: (2x^2 - x - 12 = 0) - Here, (a = 2), (b = -1), and (c = -12). - Constant term (c) is -12.- Option C: (2(x^2 - 1) = 3(x - 1)) - Expand and simplify: - Left side: (2x^2 - 2) - Right side: (3x - 3) - Bring all terms to the left: (2x^2 - 2 - 3x + 3 = 0) - Simplified: (2x^2 - 3x + 1 = 0) - Here, (a = 2), (b = -3), and (c = 1). - Constant term (c) is 1.- Option D: (2(x^2 + 1) = x + 4) - Expand and simplify: - Left side: (2x^2 + 2) - Right side: (x + 4) - Bring all terms to the left: (2x^2 + 2 - x - 4 = 0) - Simplified: (2x^2 - x - 2 = 0) - Here, (a = 2), (b = -1), and (c = -2). - Constant term (c) is -2.The only equation with a constant term of zero is Option A.(boxed{A})]

🔑:A: In x^2+x=0, the constant term c=0. B: For 2x^2-x-12=0, the constant c=-12. C: After rearranging 2(x^2-1)=3(x-1), we get 2x^2-3x+1=0, with the constant c=1. D: After rearranging, we get 2x^2-x-2=0, with the constant term c=-2. Therefore, the correct choice is: boxed{text{A}} Analysis: According to the quadratic equation ax^2+bx+c=0, the constant term is c. We can verify this by examining each option separately.