🧠:[<think>Okay, let's see. The question is about the exterior angle sum of a polygon as the number of sides increases. The options are whether it increases, decreases, remains the same, or can't be determined. Hmm. I remember something about polygons and their angles from geometry class. Let me try to recall.First, I think there's a formula for the sum of exterior angles of a polygon. If I remember correctly, for any convex polygon, the sum of the exterior angles is always 360 degrees. Wait, is that right? Let me check. So, for a triangle, each exterior angle is 120 degrees, but three times 120 is 360. For a square, each exterior angle is 90 degrees, four sides, so 4 times 90 is 360. Oh, so maybe regardless of the number of sides, the sum of the exterior angles is always 360 degrees? That seems familiar. So if that's the case, then even if the number of sides increases, the sum remains 360 degrees. So the answer would be C: Remains the same.But wait, let me make sure I'm not confusing exterior angles with interior angles. The interior angles do change with the number of sides. For example, a triangle's interior angles sum to 180 degrees, a quadrilateral's sum to 360, pentagon 540, and so on. The formula for the sum of interior angles is (n-2)*180, where n is the number of sides. So as n increases, the sum of interior angles increases. But the question is about exterior angles. Right, each exterior angle is supplementary to the interior angle. So for a regular polygon, each exterior angle would be 180 - interior angle. But when considering the sum of all exterior angles, I think regardless of the polygon being regular or not, the sum is always 360 degrees. Is that true? Let me test with a convex polygon. If you walk around the polygon, making one full rotation, the total turn would be 360 degrees. That's kind of the intuition. Each exterior angle represents the angle you turn at each vertex when walking around the polygon. So after completing the polygon, you've turned a full circle, which is 360 degrees. So regardless of the number of sides, the total turn remains 360. Therefore, the sum of exterior angles is always 360. So even if the polygon has more sides, the sum stays the same. Therefore, the answer should be C: Remains the same.But let me check for another shape. Let's take a regular hexagon. Each exterior angle would be 60 degrees, since 360/6 = 60. So sum is 6*60 = 360. Then a heptagon, 7 sides, each exterior angle is 360/7 ≈ 51.43 degrees. Sum is 7*(360/7) = 360. Yeah, so it holds. So even as n increases, each individual exterior angle gets smaller, but their total sum remains 360. So even if it's a polygon with 100 sides, each exterior angle is 3.6 degrees, and sum is 100*3.6 = 360. So that checks out. So the sum remains the same, regardless of the number of sides. Therefore, the correct answer is C.Final Answerboxed{C}</think>As the number of sides of a polygon increases, its exterior angle sum remains the same. To determine this, recall that for any convex polygon, the sum of the exterior angles is always 360 degrees. This is because each exterior angle represents the angle turned at each vertex when traversing the polygon, resulting in a full rotation of 360 degrees. For example:- A triangle has three exterior angles, each of 120 degrees, summing to (3 times 120 = 360) degrees.- A square has four exterior angles, each of 90 degrees, summing to (4 times 90 = 360) degrees.- A regular hexagon has six exterior angles, each of 60 degrees, summing to (6 times 60 = 360) degrees.This pattern holds for any polygon, regardless of the number of sides. As the number of sides (n) increases, each individual exterior angle becomes smaller, but their total sum remains 360 degrees. Thus, the exterior angle sum of a polygon remains the same as the number of sides increases.[boxed{C}],<think>Okay, so the question here is about polygons and their exterior angle sums. The question asks what happens to the exterior angle sum as the number of sides of a polygon increases. The options are A: Increases, B: Decreases, C: Remains the same, D: Cannot be determined.Hmm, let me recall what I know about polygon angles. I remember that for any convex polygon, the sum of the exterior angles is always 360 degrees. Wait, is that right? Let me think. So, if you take a regular polygon, like a triangle, square, pentagon, etc., each exterior angle is calculated as 360 divided by the number of sides. So, for a triangle, each exterior angle is 120 degrees, but there are three of them, so 3 times 120 is 360. For a square, each exterior angle is 90 degrees, four of them, so 4 times 90 is still 360. So yeah, the sum of all exterior angles of any convex polygon is always 360 degrees. So that's regardless of the number of sides. Therefore, if the number of sides increases, the measure of each exterior angle decreases (since 360/n gets smaller as n increases), but the sum remains 360 degrees. So the sum doesn't change. Therefore, the answer should be C: Remains the same.Wait, but the question is about the exterior angle sum. So even if the number of sides increases, the total sum of all exterior angles is still 360. Therefore, whether it's a triangle, quadrilateral, pentagon, etc., the sum is always 360. So the answer is C.But let me double-check. Maybe I'm confusing something. Suppose we have a polygon with more sides, say a decagon (10 sides). Each exterior angle is 36 degrees, so 10 times 36 is 360. Yeah, that's the same. So regardless of the number of sides, the sum is 360. Therefore, increasing the number of sides doesn't affect the total sum. So the answer is C. The sum remains the same.Therefore, the correct answer is C.Final Answerboxed{C}</think>As the number of sides of a polygon increases, the sum of its exterior angles remains the same. This is because, for any convex polygon, the sum of the exterior angles is always (360^circ), regardless of the number of sides. Each exterior angle of a regular polygon is calculated as (frac{360^circ}{n}), where (n) is the number of sides. While the measure of each individual exterior angle decreases as the number of sides increases, the total sum of all exterior angles remains (360^circ). Thus, the correct answer is (boxed{C}).]

🔑:AnalysisThis question mainly tests the understanding of the theorem regarding the sum of the exterior angles of a polygon. Correctly understanding this theorem is key. The sum of the exterior angles of any polygon is 360 degrees, which is independent of the number of sides.AnswerSolution: If the number of sides of a polygon increases, its exterior angle sum is 360^circ, which remains the same. Therefore, the correct answer is boxed{C}.

🧠:[<think>Alright, let's tackle this problem step by step. The question is asking for the sum of the real roots of the equation ( x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ). The answer choices are A through E, with numbers 1 to 5. Hmm, okay. First, I remember that for any polynomial equation, the sum of all its roots (real and complex) can be found using Vieta's formulas. For a quartic equation ( x^4 + ax^3 + bx^2 + cx + d = 0 ), the sum of the roots is ( -a ). In this case, the equation is ( x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ), so the coefficient ( a ) is -4. Therefore, the sum of all roots is ( -(-4) = 4 ). But wait, the question specifically asks for the sum of the real roots. So, if there are complex roots, they might come in conjugate pairs, and their sum would also be a real number. However, since complex roots come in pairs, their sum is 2 times the real part of each root. But I need to figure out how many real roots there are and then sum them up.Looking at the equation ( x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ), it might be a palindromic polynomial. Let me check. A palindromic polynomial has coefficients that read the same forwards and backwards. Here, the coefficients are 1, -4, 5, -4, 1. Yep, that's symmetric. So, it's a palindromic polynomial of even degree (4th degree). For such polynomials, there's a substitution that can simplify the equation. Specifically, if we let ( y = x + frac{1}{x} ), but since it's a quartic, maybe dividing both sides by ( x^2 ) would help. Let me try that.Dividing both sides by ( x^2 ), we get:( x^2 - 4x + 5 - frac{4}{x} + frac{1}{x^2} = 0 )Let's rearrange terms:( x^2 + frac{1}{x^2} - 4x - frac{4}{x} + 5 = 0 )Notice that ( x^2 + frac{1}{x^2} ) can be written as ( left( x + frac{1}{x} right)^2 - 2 ), and similarly, ( -4x - frac{4}{x} = -4left( x + frac{1}{x} right) ). Let me set ( y = x + frac{1}{x} ). Then, ( x^2 + frac{1}{x^2} = y^2 - 2 ), so substituting back into the equation:( (y^2 - 2) - 4y + 5 = 0 )Simplify:( y^2 - 2 - 4y + 5 = y^2 - 4y + 3 = 0 )So, the equation reduces to ( y^2 - 4y + 3 = 0 ). Let's solve for y.Using the quadratic formula: ( y = frac{4 pm sqrt{16 - 12}}{2} = frac{4 pm 2}{2} ), which gives ( y = 3 ) or ( y = 1 ).So, we have two cases for y:1. ( x + frac{1}{x} = 3 )2. ( x + frac{1}{x} = 1 )Let's solve each equation for x.Starting with the first equation: ( x + frac{1}{x} = 3 ). Multiply both sides by x (assuming ( x neq 0 )):( x^2 + 1 = 3x )Bring all terms to one side:( x^2 - 3x + 1 = 0 )Solve using quadratic formula:( x = frac{3 pm sqrt{9 - 4}}{2} = frac{3 pm sqrt{5}}{2} )So, two solutions here: ( frac{3 + sqrt{5}}{2} ) and ( frac{3 - sqrt{5}}{2} ). Let me check if these are real numbers. Since sqrt(5) is real, both roots are real.Now, the second equation: ( x + frac{1}{x} = 1 ). Multiply both sides by x:( x^2 + 1 = x )Bring terms to one side:( x^2 - x + 1 = 0 )Quadratic formula here gives:( x = frac{1 pm sqrt{1 - 4}}{2} = frac{1 pm sqrt{-3}}{2} )So, these roots are complex: ( frac{1 + isqrt{3}}{2} ) and ( frac{1 - isqrt{3}}{2} ). So, these are complex conjugates.Therefore, the original quartic equation has two real roots and two complex roots. The real roots are ( frac{3 + sqrt{5}}{2} ) and ( frac{3 - sqrt{5}}{2} ). Let's compute their sum:( frac{3 + sqrt{5}}{2} + frac{3 - sqrt{5}}{2} = frac{3 + sqrt{5} + 3 - sqrt{5}}{2} = frac{6}{2} = 3 )Wait, so the sum is 3. But according to Vieta's formula, the sum of all roots is 4, and the sum of the complex roots is ( frac{1 + isqrt{3}}{2} + frac{1 - isqrt{3}}{2} = frac{2}{2} = 1 ). Then, 3 + 1 = 4, which matches Vieta's result. So that seems correct.But the answer choices are A)5, B)4, C)3, D)2, E)1. So the answer should be C)3.Wait, but let me double-check. Maybe I made a mistake in assuming that the substitution is valid. Let's verify the roots.Take the real roots: ( frac{3 + sqrt{5}}{2} approx frac{3 + 2.236}{2} = frac{5.236}{2} approx 2.618 ). Let's plug into the original equation:( x^4 -4x^3 +5x^2 -4x +1 )Calculating each term for x ≈ 2.618:x ≈ 2.618x^4 ≈ (2.618)^4 ≈ (2.618)^2 * (2.618)^2 ≈ 6.854 * 6.854 ≈ 46.97-4x^3 ≈ -4*(2.618)^3 ≈ -4*(17.944) ≈ -71.7765x^2 ≈ 5*(6.854) ≈ 34.27-4x ≈ -4*(2.618) ≈ -10.472+1 ≈ 1Adding all together: 46.97 -71.776 +34.27 -10.472 +1 ≈ Let's compute step by step:46.97 -71.776 = -24.806-24.806 +34.27 = 9.4649.464 -10.472 = -1.008-1.008 +1 ≈ -0.008. Hmm, close to zero, which is probably due to rounding errors. So that's approximately zero, which checks out.Similarly, for x ≈ (3 - sqrt(5))/2 ≈ (3 - 2.236)/2 ≈ 0.764/2 ≈ 0.382Let's compute the original equation at x ≈ 0.382:x^4 ≈ (0.382)^4 ≈ ≈ (0.382)^2 * (0.382)^2 ≈ 0.1459 * 0.1459 ≈ 0.0213-4x^3 ≈ -4*(0.382)^3 ≈ -4*(0.0557) ≈ -0.22285x^2 ≈ 5*(0.1459) ≈ 0.7295-4x ≈ -4*(0.382) ≈ -1.528+1 ≈ 1Adding all terms: 0.0213 -0.2228 +0.7295 -1.528 +1 ≈ 0.0213 -0.2228 ≈ -0.2015-0.2015 +0.7295 ≈ 0.5280.528 -1.528 ≈ -1.0-1.0 +1 ≈ 0.0. Again, rounding errors, but it's close to zero. So both roots check out.Therefore, the real roots are indeed those two, and their sum is 3, which is answer choice C. However, the options given are (A)5, (B)4, (C)3, (D)2, (E)1. So, the correct answer should be C)3. But wait, just to make sure that there aren't more real roots. The original equation is a quartic, so up to four real roots, but in this case, we found two real roots and two complex roots. Let me confirm by checking the graph of the polynomial.Alternatively, maybe I can check the derivative to see the number of real roots. Let's compute the derivative:f(x) = x^4 -4x^3 +5x^2 -4x +1f'(x) = 4x^3 -12x^2 +10x -4To find critical points, set f'(x) = 0. Solving 4x^3 -12x^2 +10x -4 = 0. Hmm, this might be complicated, but maybe we can factor it.Let me try rational roots. Possible rational roots are ±1, ±2, ±4, ±1/2, etc.Test x=1: 4 -12 +10 -4 = -2 ≠0x=2: 32 -48 +20 -4 = 0. So x=2 is a root.So, factor out (x -2):Using polynomial division or synthetic division.Divide 4x^3 -12x^2 +10x -4 by (x -2):Coefficients: 4 | -12 | 10 | -4Bring down 4. Multiply by 2: 8. Add to -12: -4Multiply by 2: -8. Add to 10: 2Multiply by 2: 4. Add to -4: 0. So the quotient is 4x^2 -4x +2.Therefore, f'(x) = (x -2)(4x^2 -4x +2). Now set 4x^2 -4x +2 =0.Discriminant: 16 - 32 = -16 <0. So, the derivative has only one real critical point at x=2. Therefore, the original function has a critical point at x=2, and since the leading coefficient is positive, the function tends to +infinity as x approaches ±infinity. Let's check the behavior around x=2.Compute f(2): 16 -32 +20 -8 +1 = -3. So f(2) = -3.Since there is only one critical point, the function decreases to x=2, reaching a minimum at x=2 (since the derivative changes from negative to positive?), wait. Wait, f'(x) is 0 at x=2. Let me check the sign of the derivative around x=2.For x <2, say x=1: f'(1)=4 -12 +10 -4= -2 <0For x >2, say x=3: f'(3)=4*27 -12*9 +10*3 -4=108 -108 +30 -4=26>0Therefore, the function decreases before x=2 and increases after x=2. Thus, x=2 is a local minimum. Since f(2)=-3, and the function approaches +infinity as x approaches both ±infinity, this suggests that the graph crosses the x-axis twice: once to the left of x=2 and once to the right. But wait, when x approaches negative infinity, x^4 dominates, so it goes to +infinity, and near x=0, let's compute f(0)=0 -0 +0 -0 +1=1>0. At x=2, f(2)=-3, so between x=2 and infinity, the function goes from -3 to +infinity, so it must cross the x-axis once there. Similarly, left of x=2, from x=-infty to x=2, the function goes from +infinity to -3, so must cross the x-axis once. However, we already found two real roots, both positive. Wait, this seems contradictory. Wait, the roots we found are approximately 2.618 and 0.382. So, both positive. But according to the derivative analysis, there should be a root to the left of x=2 and another to the right. But 0.382 is to the left of x=2, and 2.618 is to the right. So that matches. Wait, but in the left of x=2, as x approaches -infty, the function is positive, but there's a root at x≈0.382, which is positive. Wait, perhaps there's a confusion here. Wait, in the interval from x= -infty to x=2, the function goes from +infty to -3, but if it's positive at x=0 (f(0)=1) and negative at x=2 (f(2)=-3), that means it crosses the x-axis between x=0 and x=2. Similarly, between x=2 and +infty, it goes from -3 to +infty, crossing once. So that's two real roots, which matches our previous conclusion. So, the real roots are both positive: one between 0 and 2, and another greater than 2. Which are approximately 0.382 and 2.618. So their sum is 3. So the answer is C.Therefore, after careful consideration, I believe the answer is C)3.Final Answerboxed{C}</think>To find the sum of the real roots of the equation ( x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ), we start by recognizing that it is a palindromic polynomial. Dividing by ( x^2 ) and substituting ( y = x + frac{1}{x} ), we transform the equation:[x^2 + frac{1}{x^2} - 4left( x + frac{1}{x} right) + 5 = 0]This simplifies to:[(y^2 - 2) - 4y + 5 = 0 implies y^2 - 4y + 3 = 0]Solving the quadratic equation ( y^2 - 4y + 3 = 0 ), we find the roots ( y = 3 ) and ( y = 1 ).For ( y = 3 ), we solve ( x + frac{1}{x} = 3 ):[x^2 - 3x + 1 = 0 implies x = frac{3 pm sqrt{5}}{2}]These roots are real: ( frac{3 + sqrt{5}}{2} ) and ( frac{3 - sqrt{5}}{2} ).For ( y = 1 ), we solve ( x + frac{1}{x} = 1 ):[x^2 - x + 1 = 0 implies x = frac{1 pm isqrt{3}}{2}]These roots are complex.The real roots are ( frac{3 + sqrt{5}}{2} ) and ( frac{3 - sqrt{5}}{2} ). Their sum is:[frac{3 + sqrt{5}}{2} + frac{3 - sqrt{5}}{2} = frac{6}{2} = 3]Thus, the sum of the real roots is (boxed{C}).,<think>Okay, so I have this equation: x⁴ - 4x³ + 5x² - 4x + 1 = 0. And I need to find the sum of the real roots. The answer choices are from A to E, with numbers 1 to 5. Hmm, let's see. First, I remember that for any polynomial equation, the sum of all roots (real and complex) is given by the coefficient of the term with degree one less than the leading term, divided by the leading coefficient. Since this is a quartic equation (degree 4), the sum of all roots should be 4 (since the coefficient of x³ is -4, and dividing by the leading coefficient 1 gives -(-4)/1 = 4). But wait, the question specifically asks for the sum of real roots. So, even though the sum of all roots is 4, some of them might be complex, which we don't want to include. So I need to find how many real roots there are and add them up.But how do I find the real roots? Quartic equations can be tricky, but maybe this one factors nicely. Let me try to factor it. Let me see. The equation is x⁴ - 4x³ + 5x² - 4x + 1 = 0. Maybe it's a quadratic in terms of x²? Let me check. If I let y = x², then the equation becomes y² - 4x³ + 5y - 4x + 1. Hmm, but there's still x³ and x terms. That doesn't help. Maybe substitution. Sometimes, symmetric equations can be solved by substitution. Let's look at the coefficients: 1, -4, 5, -4, 1. Notice that the coefficients are symmetric: 1, -4, 5, -4, 1. Wait, is that true? Let's check: first term is 1, then -4, 5, -4, 1. Yes, the coefficients are palindromic. That is, they read the same forwards and backwards. For palindromic polynomials, there is a property that if x is a root, then 1/x is also a root. Let me verify that.If we substitute x = 1/x into the polynomial, we get:(1/x)⁴ - 4(1/x)³ + 5(1/x)² - 4(1/x) + 1.Multiply both sides by x⁴ to clear the denominators:1 - 4x + 5x² - 4x³ + x⁴.Which is the same as the original polynomial. So yes, it is a reciprocal polynomial. Therefore, if x is a root, then 1/x is also a root. So the roots come in reciprocal pairs. That means that if r is a root, then 1/r is also a root. However, this doesn't necessarily mean they are real. If r is real, then 1/r is real. If r is complex, then 1/r is also complex. So, perhaps this polynomial can be factored into quadratic terms, each of which corresponds to a reciprocal pair.Let me try to factor it into quadratics. Let's suppose that x⁴ -4x³ +5x² -4x +1 factors into (x² + ax + 1)(x² + bx + 1). Let me check if this is possible. Multiply these two quadratics:(x² + ax + 1)(x² + bx + 1) = x⁴ + (a + b)x³ + (ab + 2)x² + (a + b)x + 1.Compare this to the original equation x⁴ -4x³ +5x² -4x +1. Therefore, we need:a + b = -4 (from the x³ term)ab + 2 = 5 (from the x² term)And also, the coefficients of x: a + b = -4 (same as the x³ term). That's consistent.So, solving the system:a + b = -4ab = 5 - 2 = 3.So we have two numbers a and b such that their sum is -4 and product is 3. Let's solve for a and b. Let me set up the quadratic equation: t² +4t +3 = 0. Factoring: (t +1)(t +3) = 0. So t = -1 or t = -3. Therefore, a = -1, b = -3 or vice versa.Therefore, the polynomial factors into (x² - x + 1)(x² -3x +1). Let's verify this:First, multiply (x² - x + 1)(x² -3x +1):= x²(x² -3x +1) -x(x² -3x +1) +1(x² -3x +1)= x⁴ -3x³ +x² -x³ +3x² -x +x² -3x +1Combine like terms:x⁴ + (-3x³ -x³) + (x² +3x² +x²) + (-x -3x) +1= x⁴ -4x³ +5x² -4x +1Yes! That's exactly the original polynomial. Great, so the factorization is correct.Therefore, the equation factors into (x² -x +1)(x² -3x +1) =0. Therefore, the roots are the roots of each quadratic.First quadratic: x² -x +1 =0. Let's find its roots. The discriminant is b² -4ac = (-1)² -4(1)(1) =1 -4= -3. Since the discriminant is negative, this quadratic has two complex conjugate roots. So, no real roots here.Second quadratic: x² -3x +1 =0. Let's compute the discriminant: (-3)² -4(1)(1)=9 -4=5. Since the discriminant is positive, this quadratic has two distinct real roots. Therefore, the original equation has two real roots, which are the roots of x² -3x +1=0.Now, the sum of the real roots is equal to the sum of the roots of x² -3x +1=0. For a quadratic equation ax² +bx +c=0, the sum of roots is -b/a. Here, a=1, b=-3, so sum is -(-3)/1=3. Therefore, the sum of real roots is 3. But wait, the answer choices are A)5, B)4, C)3, D)2, E)1. So, 3 is option C. Hmm. But wait, before I conclude, let me double-check.Wait, so the original polynomial is quartic, so it has four roots. The two quadratics each have two roots. The first quadratic (x² -x +1) has two complex roots, and the second quadratic (x² -3x +1) has two real roots. Therefore, all real roots come from the second quadratic, and their sum is 3. But the sum of all roots (real and complex) is 4, as per Vieta's formula. So, the sum of complex roots would be 4 -3=1. But complex roots come in conjugate pairs, so their sum should be 2 times the real part. If the roots are reciprocal, then complex roots would be reciprocal pairs. Wait, but for the first quadratic x² -x +1=0, the roots are reciprocal? Let's check. If r is a root, then 1/r is also a root. Let me see. For the equation x² -x +1=0, roots are (1 ± i√3)/2. Then 1/r would be (2)/(1 ± i√3) = (2)(1 ∓ i√3)/(1 + 3) = (1 ∓ i√3)/2. Which is the other root. So yes, they are reciprocal. So the complex roots are reciprocal pairs, as expected. So their product is 1, which is the constant term. So, sum of complex roots is sum of (1 + i√3)/2 and (1 - i√3)/2, which is 1. So 1. So total sum is 3 +1=4, which matches Vieta. So the sum of real roots is 3, which is option C. Therefore, the answer should be C)3.Wait, but let me check again. Maybe I made a mistake in the factorization step. Let me verify the factorization once more. (x² -x +1)(x² -3x +1) =x⁴ -3x³ +x² -x³ +3x² -x +x² -3x +1. Then combining terms: x⁴ -4x³ +5x² -4x +1. That's correct. So, the factorization is correct. Therefore, the sum of real roots is 3. So answer is C.But just to be thorough, maybe I can compute the real roots explicitly and add them. For the quadratic x² -3x +1=0, the roots are [3 ±√(9 -4)]/2 = [3 ±√5]/2. So the two roots are (3 +√5)/2 and (3 -√5)/2. Adding them: (3 +√5)/2 + (3 -√5)/2 = (3 +3)/2 =6/2=3. So yes, the sum is indeed 3. So answer is C.Final Answerboxed{C}</think>The given equation is (x^4 - 4x^3 + 5x^2 - 4x + 1 = 0). To find the sum of the real roots, we first note that the polynomial is palindromic, meaning it reads the same forwards and backwards. This implies that if (x) is a root, then (1/x) is also a root.We factor the polynomial by recognizing it can be written as a product of two quadratic polynomials. We assume the factorization ((x^2 + ax + 1)(x^2 + bx + 1)) and solve for (a) and (b):1. Expanding ((x^2 + ax + 1)(x^2 + bx + 1)) gives (x^4 + (a + b)x^3 + (ab + 2)x^2 + (a + b)x + 1).2. Comparing coefficients with the original polynomial, we get: - (a + b = -4) - (ab + 2 = 5)Solving these equations, we find (a = -1) and (b = -3) (or vice versa). Thus, the polynomial factors as ((x^2 - x + 1)(x^2 - 3x + 1)).Next, we analyze the roots of each quadratic factor:- The quadratic (x^2 - x + 1 = 0) has a discriminant of ((-1)^2 - 4 cdot 1 cdot 1 = -3), indicating complex roots.- The quadratic (x^2 - 3x + 1 = 0) has a discriminant of ((-3)^2 - 4 cdot 1 cdot 1 = 5), indicating real roots.The roots of (x^2 - 3x + 1 = 0) are (frac{3 + sqrt{5}}{2}) and (frac{3 - sqrt{5}}{2}). Summing these roots gives:[frac{3 + sqrt{5}}{2} + frac{3 - sqrt{5}}{2} = frac{3 + 3}{2} = 3]Thus, the sum of the real roots is (boxed{C}).]

🔑:1. Given the polynomial equation: [ x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ] We need to find the sum of the real roots of this equation.2. Let's try to factorize the polynomial. Notice that the polynomial can be rewritten as: [ (x-1)^4 - x^2 = 0 ]3. Expanding ((x-1)^4): [ (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 ] Therefore, the equation becomes: [ x^4 - 4x^3 + 6x^2 - 4x + 1 - x^2 = 0 ] Simplifying, we get: [ x^4 - 4x^3 + 5x^2 - 4x + 1 = 0 ]4. Now, we factorize the polynomial: [ (x-1)^4 - x^2 = 0 ] This can be rewritten as: [ (x-1)^4 = x^2 ] Taking the square root on both sides: [ (x-1)^2 = pm x ]5. Solving for ( (x-1)^2 = x ): [ (x-1)^2 = x ] [ x^2 - 2x + 1 = x ] [ x^2 - 3x + 1 = 0 ] Using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ x = frac{3 pm sqrt{9 - 4}}{2} = frac{3 pm sqrt{5}}{2} ] These are the real roots from this factor.6. Solving for ( (x-1)^2 = -x ): [ (x-1)^2 = -x ] [ x^2 - 2x + 1 = -x ] [ x^2 - x + 1 = 0 ] The discriminant of this quadratic equation is: [ Delta = (-1)^2 - 4 cdot 1 cdot 1 = 1 - 4 = -3 ] Since the discriminant is negative, this quadratic equation has no real roots.7. Therefore, the only real roots are from the equation ( x^2 - 3x + 1 = 0 ), which are: [ x = frac{3 + sqrt{5}}{2} quad text{and} quad x = frac{3 - sqrt{5}}{2} ]8. The sum of these real roots is: [ frac{3 + sqrt{5}}{2} + frac{3 - sqrt{5}}{2} = frac{(3 + sqrt{5}) + (3 - sqrt{5})}{2} = frac{6}{2} = 3 ]Conclusion:[boxed{3}]

🧠:[<think>Okay, let's see. The problem is asking for the range of the expression ( s = frac{a+b}{1+c} + frac{b+c}{1+a} + frac{c+a}{1+b} ), where ( a, b, c ) are each in the interval ([ frac{1}{2}, 1 ]). Hmm. So I need to find the minimum and maximum possible values of ( s ) given that ( a, b, c ) can vary between 0.5 and 1. First, maybe I should try to understand the structure of this expression. Each term in the sum is a fraction where the numerator is the sum of two variables, and the denominator is 1 plus the third variable. For example, the first term is ( frac{a + b}{1 + c} ), then similarly cyclically permuted for the other terms. Since all variables are between 0.5 and 1, the denominators ( 1 + c ), ( 1 + a ), ( 1 + b ) will be between ( 1 + 0.5 = 1.5 ) and ( 1 + 1 = 2 ). Similarly, the numerators ( a + b ), ( b + c ), ( c + a ) will be between ( 0.5 + 0.5 = 1 ) and ( 1 + 1 = 2 ). So each individual fraction is between ( 1/2 ) and ( 2/1.5 approx 1.333... ). But since there are three such terms, adding them up might lead to a total range from maybe ( 3*(1/2) = 1.5 ) to ( 3*(2/1.5) = 4 ). But that's a very rough estimate. The actual range is probably narrower because the variables are linked; changing one variable affects two terms in the sum. To find the exact range, I need to analyze how ( s ) changes as ( a, b, c ) vary. Since all variables are symmetric in the expression, maybe the maximum and minimum occur when variables are at their extremes, either all equal, or some combination of 0.5 and 1. Let's test some specific cases.First, when all variables are equal. Let’s set ( a = b = c = frac{1}{2} ). Then each numerator ( a + b = 0.5 + 0.5 = 1 ), and each denominator ( 1 + c = 1 + 0.5 = 1.5 ). So each term is ( 1 / 1.5 = 2/3 ), so the total ( s = 3*(2/3) = 2 ). If all variables are equal to 1, then each numerator is ( 1 + 1 = 2 ), and each denominator is ( 1 + 1 = 2 ), so each term is 1. Thus, ( s = 3*1 = 3 ). So when all variables are equal, the value of ( s ) is between 2 and 3. But maybe when variables are unequal, the sum can be higher or lower.Let me try setting two variables to 1 and one to 0.5. Let’s say ( a = b = 1 ), ( c = 0.5 ). Then compute each term:First term: ( frac{a + b}{1 + c} = frac{1 + 1}{1 + 0.5} = frac{2}{1.5} approx 1.333... )Second term: ( frac{b + c}{1 + a} = frac{1 + 0.5}{1 + 1} = frac{1.5}{2} = 0.75 )Third term: ( frac{c + a}{1 + b} = frac{0.5 + 1}{1 + 1} = frac{1.5}{2} = 0.75 )Adding these up: ( 1.333... + 0.75 + 0.75 ≈ 2.833... ), which is approximately 2.833, which is less than 3. Hmm, so even when two variables are at maximum and one at minimum, the sum is still lower than when all variables are at maximum. So maybe the maximum is indeed 3 when all variables are 1.What about the minimum? Let's try setting two variables to 0.5 and one to 1. Let's say ( a = b = 0.5 ), ( c = 1 ). Then:First term: ( frac{0.5 + 0.5}{1 + 1} = frac{1}{2} = 0.5 )Second term: ( frac{0.5 + 1}{1 + 0.5} = frac{1.5}{1.5} = 1 )Third term: ( frac{1 + 0.5}{1 + 0.5} = frac{1.5}{1.5} = 1 )Total ( s = 0.5 + 1 + 1 = 2.5 ). That's higher than the case when all variables are 0.5, which gave ( s = 2 ). Wait, so setting two variables to 0.5 and one to 1 gives a higher sum than when all are 0.5. So maybe the minimum is when all variables are 0.5? Wait, but earlier when all variables were 0.5, we had ( s = 2 ). Let me check that again.Wait, no: if ( a = b = c = 0.5 ), then each numerator is ( 0.5 + 0.5 = 1 ), and denominator is ( 1 + 0.5 = 1.5 ), so each term is ( 1 / 1.5 = 2/3 approx 0.666... ), so total ( s = 3*(2/3) = 2 ). But when two variables are 0.5 and one is 1, we get 2.5, which is higher. So maybe the minimum is 2. But wait, is there a case where ( s ) can be lower than 2? Let's see. Let's try setting one variable to 0.5 and others to something else. For example, let’s set ( a = 0.5 ), ( b = 0.5 ), ( c = 0.5 ). That gives 2, as above. If we set two variables to 0.5 and the third to something higher, say 0.5, but that's the same as all 0.5. If we set one variable to 0.5 and others to 1, that gives higher s. So maybe 2 is indeed the minimum.But let's test another case. Suppose ( a = 0.5 ), ( b = 1 ), ( c = 0.5 ). Then:First term: ( (0.5 + 1)/(1 + 0.5) = 1.5/1.5 = 1 )Second term: ( (1 + 0.5)/(1 + 0.5) = 1.5/1.5 = 1 )Third term: ( (0.5 + 0.5)/(1 + 1) = 1/2 = 0.5 )Total ( s = 1 + 1 + 0.5 = 2.5 ). Same as before. So again, higher than 2. Alternatively, suppose we set ( a = 0.5 ), ( b = 0.6 ), ( c = 0.5 ). Maybe not all at extremes. Let's compute:First term: ( (0.5 + 0.6)/(1 + 0.5) = 1.1/1.5 ≈ 0.733 )Second term: ( (0.6 + 0.5)/(1 + 0.5) = 1.1/1.5 ≈ 0.733 )Third term: ( (0.5 + 0.5)/(1 + 0.6) = 1/1.6 ≈ 0.625 )Total ( s ≈ 0.733 + 0.733 + 0.625 ≈ 2.091 ). So that's about 2.091, which is still above 2. So even with some variables not at extremes, the sum is still higher than 2.Alternatively, let's set all variables slightly above 0.5. Let's say ( a = b = c = 0.6 ).Each term: ( (0.6 + 0.6)/(1 + 0.6) = 1.2/1.6 = 0.75 )So each term is 0.75, total ( s = 3*0.75 = 2.25 ). Still higher than 2.Wait, but when all variables are 0.5, s is 2. So maybe the minimal value is indeed 2, achieved when all variables are 0.5. Then the maximum is 3, when all variables are 1. But wait, earlier when two variables were 1 and one was 0.5, s was approximately 2.833. So 3 is higher. So the maximum is 3, minimum is 2. Is that the case? Let me check another combination.Suppose we set one variable to 1, and the other two to 0.5. Then as above, s is 2.5. If we set two variables to 1 and one to 0.5, s is approx 2.833. If we set all variables to 1, s is 3. So as variables increase, s increases. Similarly, as variables decrease, s decreases. So perhaps the minimal and maximal are at the extremes when all variables are 0.5 or all 1. That would make sense because if you have variables in the denominators, increasing a variable in the denominator would decrease the term, but increasing variables in the numerators would increase the term. So there's a balance. But since all variables are in both numerators and denominators, maybe the overall effect is that increasing all variables leads to an increase in s.Wait, let's analyze the effect of increasing a variable. Let's take variable a. If I increase a, how does that affect s? Let's see. In the term ( frac{a + b}{1 + c} ), increasing a increases the numerator. In the term ( frac{b + c}{1 + a} ), increasing a increases the denominator, thus decreasing the term. In the term ( frac{c + a}{1 + b} ), increasing a increases the numerator. So the overall effect is a combination of some terms increasing and some decreasing. So it's not straightforward. Therefore, maybe the function is not monotonic in each variable, so the extremum might not necessarily be at the corners. Therefore, maybe we need a more careful analysis.Alternatively, perhaps using calculus. Let's suppose we can take partial derivatives and set them to zero. However, with three variables, it's a bit complicated, but maybe we can assume symmetry. Suppose all variables are equal. Let ( a = b = c = t ), with ( t in [0.5, 1] ). Then the expression becomes:( s = 3 * frac{2t}{1 + t} )So ( s = frac{6t}{1 + t} ). Let's see how this function behaves as t increases from 0.5 to 1. At t = 0.5, s = 6*(0.5)/(1.5) = 3/1.5 = 2. At t = 1, s = 6*1 / 2 = 3. And the function is increasing, since derivative of ( 6t/(1 + t) ) is ( 6(1 + t) - 6t)/(1 + t)^2 = 6/(1 + t)^2 > 0 ). So when variables are equal, s increases from 2 to 3 as t increases from 0.5 to 1. So that's consistent with the earlier results.But the problem is, maybe when variables are unequal, we can get a lower or higher value. For example, when we have two variables high and one low, or vice versa.Wait, when we set two variables to 1 and one to 0.5, we got s ≈ 2.833, which is less than 3. So even though we have two variables at maximum, the third variable being low brings the sum down. Similarly, setting two variables to 0.5 and one to 1 gives s = 2.5, which is higher than the minimum 2, but lower than 3.But perhaps there's a different configuration where s is higher than 3 or lower than 2? Let's test.Suppose we set one variable to 1 and the others to something else. Wait, but variables can't exceed 1. So the maximum any variable can be is 1. Similarly, the minimum is 0.5. So maybe the maximum occurs when all variables are 1, giving s = 3. And the minimum occurs when all variables are 0.5, giving s = 2. But in the earlier test with two variables at 1 and one at 0.5, s was 2.833, which is less than 3. Similarly, two variables at 0.5 and one at 1 gives 2.5. So perhaps indeed 2 and 3 are the bounds. But let's see another case. What if we set two variables to 1, and one variable to something less than 1 but higher than 0.5? Let's say ( a = b = 1 ), ( c = 0.6 ).First term: ( (1 + 1)/(1 + 0.6) = 2/1.6 = 1.25 )Second term: ( (1 + 0.6)/(1 + 1) = 1.6/2 = 0.8 )Third term: ( (0.6 + 1)/(1 + 1) = 1.6/2 = 0.8 )Total ( s = 1.25 + 0.8 + 0.8 = 2.85 ), which is still less than 3. So even if the third variable is increased from 0.5 to 0.6, the sum increases slightly but still remains below 3. Hence, the maximum seems to occur when all variables are 1.Similarly, for the minimum, if we set two variables to 0.5 and one variable to something higher, like 0.6:( a = 0.5 ), ( b = 0.5 ), ( c = 0.6 )First term: ( (0.5 + 0.5)/(1 + 0.6) = 1/1.6 ≈ 0.625 )Second term: ( (0.5 + 0.6)/(1 + 0.5) = 1.1/1.5 ≈ 0.733 )Third term: ( (0.6 + 0.5)/(1 + 0.5) = 1.1/1.5 ≈ 0.733 )Total ( s ≈ 0.625 + 0.733 + 0.733 ≈ 2.091 ), which is still higher than 2. So even if we don't set all variables to 0.5, the sum doesn't go below 2. But maybe there's a more optimal configuration? Let's consider setting two variables to 0.5 and one variable to 0.5 as well. Wait, that's all 0.5, which gives 2. So perhaps the minimum is indeed 2. Alternatively, maybe we can use some inequality to bound the expression. Let me think. Since all variables are between 0.5 and 1, maybe we can use the AM-GM inequality or Cauchy-Schwarz. But the expression is a sum of fractions, which makes it a bit tricky.Alternatively, let's try to bound each term. For each term ( frac{a + b}{1 + c} ). Since ( a, b geq 0.5 ), the numerator ( a + b geq 1 ). The denominator ( 1 + c leq 2 ), so each term is at least ( 1/2 ). Similarly, numerator ( a + b leq 2 ), denominator ( 1 + c geq 1.5 ), so each term is at most ( 2/1.5 ≈ 1.333 ). So each term is between 0.5 and 1.333. But again, adding three such terms, the total would be between 1.5 and 4. But our earlier tests show that the actual range is 2 to 3, which is narrower. So maybe the individual term bounds are too loose. Alternatively, consider that if we can write the entire expression in terms of variables and then find its extrema. Let's attempt to do that. Let’s denote ( a, b, c in [0.5, 1] ). Then, to find the extrema of ( s = frac{a + b}{1 + c} + frac{b + c}{1 + a} + frac{c + a}{1 + b} ).Since the variables are symmetric, maybe the extrema occur at symmetric points or at the corners of the domain. We saw that at the symmetric point where all variables are equal, the expression ranges from 2 to 3. But when variables are unequal, we can get values in between. However, in our tests, the extremes (2 and 3) only occurred when all variables were equal. So perhaps those are indeed the global minima and maxima. But to confirm this, maybe we can use the method of Lagrange multipliers. Let's consider the function ( s(a, b, c) = frac{a + b}{1 + c} + frac{b + c}{1 + a} + frac{c + a}{1 + b} ), with ( a, b, c in [0.5, 1] ). The extrema can occur either at critical points inside the domain or on the boundary.First, check the critical points. To find critical points, we need to compute partial derivatives and set them to zero. Let's compute the partial derivative of ( s ) with respect to ( a ):( frac{partial s}{partial a} = frac{1}{1 + c} - frac{b + c}{(1 + a)^2} + frac{1}{1 + b} )Similarly, the partial derivatives with respect to ( b ) and ( c ):( frac{partial s}{partial b} = frac{1}{1 + c} - frac{a + c}{(1 + b)^2} + frac{1}{1 + a} )( frac{partial s}{partial c} = frac{1}{1 + a} - frac{a + b}{(1 + c)^2} + frac{1}{1 + b} )Set each derivative equal to zero. However, solving these equations simultaneously seems complex. Maybe assuming symmetry here. Suppose that at the critical point, ( a = b = c = t ). Then, substitute into the partial derivatives:For ( frac{partial s}{partial a} ):( frac{1}{1 + t} - frac{2t}{(1 + t)^2} + frac{1}{1 + t} = 0 )Simplify:( frac{2}{1 + t} - frac{2t}{(1 + t)^2} = 0 )Multiply both sides by ( (1 + t)^2 ):( 2(1 + t) - 2t = 0 )Simplify:( 2 + 2t - 2t = 0 implies 2 = 0 )Which is a contradiction. Therefore, there is no critical point with ( a = b = c ). So the critical points, if any, must be asymmetric. But solving the partial derivatives for asymmetric variables is quite complicated. Alternatively, since we didn't find any critical points inside the domain with all variables equal, maybe all extrema occur on the boundary.The boundaries of the domain occur when at least one variable is at 0.5 or 1. Let's check these cases.Case 1: All variables are at 0.5: already gives s = 2.Case 2: All variables are at 1: gives s = 3.Case 3: Two variables at 1, one at 0.5: as checked earlier, gives s ≈ 2.833.Case 4: Two variables at 0.5, one at 1: gives s = 2.5.Case 5: One variable at 1, others varying between 0.5 and 1.But perhaps the maximum occurs when all variables are 1, and the minimum when all variables are 0.5, as those give the highest and lowest s.Alternatively, let's check if there is a way to make s lower than 2. Suppose we set variables unequally, but not at the extremes. For example, set ( a = 0.5 ), ( b = 0.5 ), ( c = x ), where x is between 0.5 and 1. Then compute s in terms of x:First term: ( frac{0.5 + 0.5}{1 + x} = frac{1}{1 + x} )Second term: ( frac{0.5 + x}{1 + 0.5} = frac{0.5 + x}{1.5} )Third term: ( frac{x + 0.5}{1 + 0.5} = frac{0.5 + x}{1.5} )Total s = ( frac{1}{1 + x} + 2*frac{0.5 + x}{1.5} )Simplify:( frac{1}{1 + x} + frac{2(0.5 + x)}{1.5} = frac{1}{1 + x} + frac{1 + 2x}{1.5} )Let’s compute this as a function of x in [0.5, 1]. Let's see how it behaves. At x = 0.5:( frac{1}{1.5} + frac{1 + 1}{1.5} = frac{1}{1.5} + frac{2}{1.5} = 3/1.5 = 2 ). At x = 1:( frac{1}{2} + frac{1 + 2}{1.5} = 0.5 + 2 = 2.5 ). Let's compute the derivative with respect to x:( d/dx [ frac{1}{1 + x} + frac{1 + 2x}{1.5} ] = - frac{1}{(1 + x)^2} + frac{2}{1.5} = - frac{1}{(1 + x)^2} + frac{4}{3} )Set derivative equal to zero:( - frac{1}{(1 + x)^2} + frac{4}{3} = 0 implies frac{4}{3} = frac{1}{(1 + x)^2} implies (1 + x)^2 = 3/4 implies 1 + x = sqrt{3}/2 approx 0.866 implies x approx -0.134 ). But x is in [0.5, 1], so this critical point is outside the domain. Therefore, the function is monotonic on [0.5,1]. Check the derivative at x = 0.5:( -1/(1.5)^2 + 4/3 ≈ -1/2.25 + 1.333 ≈ -0.444 + 1.333 ≈ 0.889 > 0 ). So the function is increasing on [0.5,1]. Therefore, the minimum in this case is 2 (at x = 0.5) and maximum 2.5 (at x =1). Thus, even when two variables are fixed at 0.5 and one varies, the minimum remains 2. So perhaps 2 is indeed the global minimum. Similarly, if we fix one variable at 1 and vary the others. Let’s set ( a = 1 ), ( b = x ), ( c = 1 ). Then compute s:First term: ( (1 + x)/(1 + 1) = (1 + x)/2 )Second term: ( (x + 1)/(1 + 1) = (x +1)/2 )Third term: ( (1 + 1)/(1 + x) = 2/(1 + x) )Total s = ( frac{1 + x}{2} + frac{x + 1}{2} + frac{2}{1 + x} = frac{2(1 + x)}{2} + frac{2}{1 + x} = (1 + x) + frac{2}{1 + x} )Let’s call this function f(x) = (1 + x) + 2/(1 + x), with x ∈ [0.5,1]. Compute f(x):At x = 0.5: f = 1.5 + 2/1.5 ≈ 1.5 + 1.333 ≈ 2.833At x =1: f = 2 + 2/2 = 2 + 1 = 3Derivative: f’(x) = 1 - 2/(1 + x)^2. Set to zero:1 - 2/(1 + x)^2 = 0 → (1 + x)^2 = 2 → 1 + x = √2 ≈ 1.414 → x ≈ 0.414. Again, outside the domain [0.5,1]. Compute derivative at x =0.5:1 - 2/(1.5)^2 ≈ 1 - 2/2.25 ≈ 1 - 0.888 ≈ 0.112 > 0. So function is increasing on [0.5,1]. Therefore, the minimum in this case is ≈2.833 and maximum 3. Thus, even when one variable is fixed at 1 and others vary, the minimum s is still higher than 2.833, which is higher than the global minimum of 2. Therefore, combining all these cases, it's clear that the minimal value of s is 2, achieved when all variables are 0.5, and the maximal value is 3, achieved when all variables are 1. But to be thorough, let's check another case where two variables are at 0.5 and one is varying. Wait, no, we already checked that. Alternatively, set two variables at 1 and one varying. Wait, similar to previous case. Let’s consider setting two variables at 1 and the third at 0.5: s ≈ 2.833, which is between 2 and 3.Is there a configuration where s exceeds 3? Let's think. If variables could exceed 1, then perhaps. But variables are capped at 1. So since all variables are at most 1, and denominators are at least 1.5, so the maximum each term can be is 2/1.5 ≈1.333, but when variables are 1, denominators are 2, so terms are 2/2 =1. Therefore, adding three 1s gives 3, which is the upper bound. Hence, s cannot exceed 3.Similarly, the minimum is 2 when all variables are 0.5, as each term is 1/1.5 ≈0.666, summing to 2. Therefore, after checking various configurations and considering the behavior of the function when variables are varied, it seems that the range of s is [2, 3]. Final AnswerThe range of ( s ) is (boxed{[2, 3]}).</think>Given ( a, b, c in left[ frac{1}{2}, 1 right] ), we need to find the range of the expression ( s = frac{a+b}{1+c} + frac{b+c}{1+a} + frac{c+a}{1+b} ).1. Testing Extreme Values: - When ( a = b = c = frac{1}{2} ): [ s = 3 cdot frac{1}{1.5} = 3 cdot frac{2}{3} = 2 ] - When ( a = b = c = 1 ): [ s = 3 cdot frac{2}{2} = 3 cdot 1 = 3 ]2. Testing Mixed Values: - Setting two variables to 1 and one to 0.5: [ text{Example: } a = b = 1, c = 0.5 implies s = frac{2}{1.5} + frac{1.5}{2} + frac{1.5}{2} approx 2.833 ] - Setting two variables to 0.5 and one to 1: [ text{Example: } a = b = 0.5, c = 1 implies s = frac{1}{2} + frac{1.5}{1.5} + frac{1.5}{1.5} = 0.5 + 1 + 1 = 2.5 ]3. Checking Intermediate Values: - Setting variables slightly above 0.5 or below 1 results in values of ( s ) between 2 and 3, but never exceeding these bounds.4. Analyzing the Function: - The function ( s ) is symmetric and each term is influenced by the variables in both the numerator and denominator. - Testing partial derivatives and critical points shows that the function does not have critical points within the domain that result in values outside the range [2, 3].5. Conclusion: - The minimum value of ( s ) is 2 when all variables are 0.5. - The maximum value of ( s ) is 3 when all variables are 1.Thus, the range of ( s ) is (boxed{[2, 3]}).,<think>Okay, so I need to find the range of the expression ( s = frac{a+b}{1+c} + frac{b+c}{1+a} + frac{c+a}{1+b} ) where ( a, b, c ) are each in the interval ([ frac{1}{2}, 1 ]). Hmm, let's think about how to approach this.First, maybe I should consider the maximum and minimum values that each term in the sum can take. Since all variables are between 0.5 and 1, the numerators and denominators will vary within certain ranges. But since each term involves two variables in the numerator and one in the denominator, it's a bit more complex. Maybe I can use some inequalities to bound the expression?Alternatively, symmetry might help here. The expression is symmetric in ( a, b, c ), so perhaps the maximum or minimum occurs when all variables are equal. Let me check that.Suppose ( a = b = c = frac{1}{2} ). Then each numerator becomes ( frac{1}{2} + frac{1}{2} = 1 ), and each denominator becomes ( 1 + frac{1}{2} = frac{3}{2} ). So each term is ( frac{1}{frac{3}{2}} = frac{2}{3} ), so total ( s = 3 times frac{2}{3} = 2 ).If all variables are 1, then each numerator is ( 1 + 1 = 2 ), denominator is ( 1 + 1 = 2 ), so each term is 1, total ( s = 3 times 1 = 3 ).So when all variables are equal, s ranges from 2 to 3. But the problem says the variables can be anywhere in [1/2, 1], not necessarily equal. So the actual range might be larger. Maybe the maximum is higher when variables are not equal, and the minimum is lower? Let's check.First, let's try to find the maximum possible value of s. To maximize each term ( frac{a + b}{1 + c} ), we would want to maximize the numerators and minimize the denominators. Since ( a, b ) can be as large as 1 and ( c ) as small as 1/2.So, if I set ( a = b = 1 ) and ( c = frac{1}{2} ), then:First term: ( frac{1 + 1}{1 + frac{1}{2}} = frac{2}{1.5} = frac{4}{3} approx 1.333 )Second term: ( frac{1 + frac{1}{2}}{1 + 1} = frac{1.5}{2} = 0.75 )Third term: ( frac{frac{1}{2} + 1}{1 + 1} = frac{1.5}{2} = 0.75 )Total ( s = frac{4}{3} + 0.75 + 0.75 = frac{4}{3} + 1.5 approx 1.333 + 1.5 = 2.833 ), which is less than 3. Wait, that's strange. When all variables are 1, s is 3. So maybe this isn't the maximum.Wait, maybe I made a mistake here. If I set two variables to 1 and one to 1/2, but the expression is symmetric, so perhaps permuting variables would give the same result. Let me check another configuration. Suppose ( a = 1 ), ( b = 1 ), ( c = 1 ). Then s is 3, as before. So perhaps 3 is indeed the maximum.But wait, maybe if two variables are 1 and one is smaller, but is 3 still the maximum? Let me recalculate. If ( a = 1 ), ( b = 1 ), ( c = 1/2 ), then:First term: ( (1 + 1)/(1 + 0.5) = 2/1.5 ≈ 1.333 )Second term: ( (1 + 0.5)/(1 + 1) = 1.5/2 = 0.75 )Third term: ( (0.5 + 1)/(1 + 1) = 1.5/2 = 0.75 )Total s ≈ 1.333 + 0.75 + 0.75 ≈ 2.833, which is less than 3. So in this case, the sum is lower. Hmm, so maybe the maximum is indeed 3 when all variables are 1. Let's see another configuration where maybe two variables are 1 and one is 1/2 but in a different position.Wait, no, because the expression is symmetric, any permutation would give the same sum. So maybe the maximum is indeed 3 when all variables are 1. Similarly, when all variables are 1/2, the sum is 2. But perhaps when variables are not equal, the sum can be lower or higher?Wait, let me test another case. Suppose two variables are 1/2 and one is 1. Let's set ( a = 1 ), ( b = 1/2 ), ( c = 1/2 ).First term: ( (1 + 0.5)/(1 + 0.5) = 1.5/1.5 = 1 )Second term: ( (0.5 + 0.5)/(1 + 1) = 1/2 = 0.5 )Third term: ( (0.5 + 1)/(1 + 0.5) = 1.5/1.5 = 1 )Total s = 1 + 0.5 + 1 = 2.5, which is between 2 and 3. So maybe the minimum is 2 when all variables are 1/2, and maximum is 3 when all variables are 1. But wait, let me check another case where one variable is 1/2 and others are 1. Wait, that's similar to the earlier case.Wait, but maybe there's a way to make s even lower than 2? Let's see. If variables can be 1/2, but maybe if some variables are higher, but others are lower? Wait, no, all variables are bounded below by 1/2. So the minimal value for each variable is 1/2. So the minimal case is when all variables are 1/2, giving s=2. But maybe if some variables are higher, others are lower, but since they can't go below 1/2, the minimal sum is when all are 1/2.Wait, but perhaps not. Maybe the interactions between numerators and denominators could lead to a lower total. Let me check.Suppose a = 1/2, b = 1/2, c = 1.First term: ( (0.5 + 0.5)/(1 + 1) = 1/2 = 0.5 )Second term: ( (0.5 + 1)/(1 + 0.5) = 1.5/1.5 = 1 )Third term: ( (1 + 0.5)/(1 + 0.5) = 1.5/1.5 = 1 )Total s = 0.5 + 1 + 1 = 2.5, which is still higher than 2. So when two variables are 1/2 and one is 1, s=2.5. So maybe indeed the minimal is 2 when all variables are 1/2.But let's check another configuration. Suppose a = 1, b = 1/2, c = 1/2. Then:First term: ( (1 + 0.5)/(1 + 0.5) = 1.5/1.5 = 1 )Second term: ( (0.5 + 0.5)/(1 + 1) = 1/2 = 0.5 )Third term: ( (0.5 + 1)/(1 + 0.5) = 1.5/1.5 = 1 )Total s = 1 + 0.5 + 1 = 2.5. Still 2.5. So s=2.5 here.Alternatively, if a = 1/2, b=1, c=1/2:First term: (0.5 + 1)/(1 + 0.5) = 1.5 / 1.5 = 1Second term: (1 + 0.5)/(1 + 0.5) = 1.5 / 1.5 = 1Third term: (0.5 + 0.5)/(1 + 1) = 1 / 2 = 0.5Total s = 1 + 1 + 0.5 = 2.5. Same as before.So, it seems that whenever two variables are 1/2 and one is 1, s=2.5. If one variable is 1/2 and others are 1, then s=2.833. If all variables are 1/2, s=2. If all variables are 1, s=3.But wait, maybe there's a way to get a sum lower than 2? Let me test a different combination. For example, set a=1/2, b=1/2, c=1/2: s=2.If a=1/2, b=1/2, c=0.5: same as above.Alternatively, suppose a=1/2, b=1/2, c=3/4. Let's compute s:First term: (0.5 + 0.5)/(1 + 0.75) = 1 / 1.75 ≈ 0.571Second term: (0.5 + 0.75)/(1 + 0.5) = 1.25 / 1.5 ≈ 0.833Third term: (0.75 + 0.5)/(1 + 0.5) = 1.25 / 1.5 ≈ 0.833Total s ≈ 0.571 + 0.833 + 0.833 ≈ 2.237, which is higher than 2 but lower than 2.5. Hmm, so maybe there's a way to get s between 2 and 2.5. But in this case, c is 0.75, so variables are not all 1/2. So perhaps the minimum is indeed 2, and as variables increase, s increases. But in this case, when c is increased to 0.75, but a and b are 0.5, s is 2.237, which is still higher than 2.Wait, so maybe if variables are not all 1/2, but some are higher, but others are lower, but they can't go below 1/2. So, perhaps to minimize s, we need to minimize each term. Each term is ( frac{a + b}{1 + c} ). To minimize this, we want to minimize the numerator and maximize the denominator. So set a and b to 1/2, and c to 1. But in this case, if two variables are 1/2 and the third is 1, then one term is (0.5 + 0.5)/(1 + 1) = 0.5, another term is (0.5 + 1)/(1 + 0.5) = 1.5/1.5 = 1, and the third term is same as the second. So total s=0.5 + 1 +1=2.5, which is higher than 2. So even if we try to minimize each term, the interaction between variables complicates things.Alternatively, maybe using calculus to find critical points? But since it's a three-variable function, it might get complicated. Maybe use Lagrange multipliers? Hmm. Let me think.Alternatively, perhaps use substitution. Since the variables are symmetric, maybe assume that a ≤ b ≤ c or something like that, but I don't know if that helps.Alternatively, consider the function when variables are at the endpoints of the interval [1/2,1], since extrema might occur at the boundaries. So maybe check all combinations where each variable is either 1/2 or 1. Let's see.There are 2^3 = 8 possible combinations where each variable is either 1/2 or 1.Let's list them:1. a=1/2, b=1/2, c=1/2: s=22. a=1/2, b=1/2, c=1: s=2.53. a=1/2, b=1, c=1/2: s=2.54. a=1, b=1/2, c=1/2: s=2.55. a=1/2, b=1, c=1: Let's compute this.First term: (0.5 + 1)/(1 + 1) = 1.5/2 = 0.75Second term: (1 + 1)/(1 + 0.5) = 2/1.5 ≈ 1.333Third term: (1 + 0.5)/(1 + 1) = 1.5/2 = 0.75Total s ≈ 0.75 + 1.333 + 0.75 ≈ 2.8336. a=1, b=1/2, c=1: Same as above, by symmetry. s≈2.8337. a=1, b=1, c=1/2: s≈2.8338. a=1, b=1, c=1: s=3So among these 8 cases, the minimal s is 2 (all 1/2), and the maximum is 3 (all 1). So perhaps the range is [2,3]. But wait, in the previous example where variables were not at the endpoints, but somewhere in between, s was 2.237. So maybe there are values of s between 2 and 3 even when variables are inside the interval. But maybe the minimal is indeed 2 and the maximal 3.But hold on, the problem says "the range of s". If we can find that s can't be less than 2 or more than 3, then the range is [2,3]. But we need to verify that for all a, b, c in [1/2,1], s is between 2 and 3.But in the earlier example where a=1/2, b=1/2, c=3/4, s was approximately 2.237, which is still between 2 and 3. So maybe 2 and 3 are indeed the bounds.Alternatively, perhaps there's a case where s can be lower than 2. Let's see. Suppose a=1/2, b=1/2, c= something.First term: (0.5 + 0.5)/(1 + c) = 1/(1 + c)Second term: (0.5 + c)/(1 + 0.5) = (0.5 + c)/1.5Third term: (c + 0.5)/(1 + 0.5) = (0.5 + c)/1.5So total s = 1/(1 + c) + 2*(0.5 + c)/1.5Let me express this as:s(c) = 1/(1 + c) + 2*(0.5 + c)/1.5Simplify:s(c) = 1/(1 + c) + (1 + 2c)/1.5Now, c is in [0.5,1]. Let's check the derivative of s(c) to see if it has a minimum in this interval.s'(c) = -1/(1 + c)^2 + (2)/1.5Set derivative to zero:-1/(1 + c)^2 + 4/3 = 0So 4/3 = 1/(1 + c)^2=> (1 + c)^2 = 3/4=> 1 + c = sqrt(3)/2 ≈ 0.866, but sqrt(3)/2 ≈ 0.866, so c ≈ -0.134, which is outside the interval [0.5,1]. Therefore, the minimum of s(c) occurs at the endpoint c=0.5 or c=1.Calculate s at c=0.5:s(0.5) = 1/(1.5) + (1 + 1)/1.5 = (2/3) + (2)/1.5 ≈ 0.666 + 1.333 = 2At c=1:s(1) = 1/2 + (1 + 2)/1.5 = 0.5 + 3/1.5 = 0.5 + 2 = 2.5Therefore, in this case, when two variables are 0.5 and one is varying between 0.5 and 1, the minimal s is 2 (at c=0.5) and maximum 2.5 (at c=1). So s can't go below 2 here.Similarly, if we fix two variables at 0.5 and vary the third, s stays between 2 and 2.5. If we fix one variable at 0.5 and others at 1, s is around 2.833. If we set all variables to 1, s=3.Alternatively, if we set two variables to 1 and one to 0.5, s is 2.833, as calculated before. So, in all cases, s seems bounded between 2 and 3.But is there a configuration where s can be less than 2? For example, if all variables are set to their minimal value 0.5, s=2. If we set variables above 0.5, then some terms may decrease but others increase. Wait, let's see. Suppose we set a=0.5, b=0.6, c=0.7. Let's compute s.First term: (0.5 + 0.6)/(1 + 0.7) = 1.1 / 1.7 ≈ 0.647Second term: (0.6 + 0.7)/(1 + 0.5) = 1.3 / 1.5 ≈ 0.867Third term: (0.7 + 0.5)/(1 + 0.6) = 1.2 / 1.6 = 0.75Total s ≈ 0.647 + 0.867 + 0.75 ≈ 2.264, which is still above 2.Another example: a=0.5, b=0.5, c=0.6. Then:First term: (0.5 + 0.5)/(1 + 0.6) = 1 / 1.6 ≈ 0.625Second term: (0.5 + 0.6)/(1 + 0.5) = 1.1 / 1.5 ≈ 0.733Third term: (0.6 + 0.5)/(1 + 0.5) = 1.1 / 1.5 ≈ 0.733Total s ≈ 0.625 + 0.733 + 0.733 ≈ 2.091, which is still above 2. So even when variables are slightly above 0.5, s is still above 2. So it seems that the minimal value is indeed 2 when all variables are 0.5.Similarly, when variables increase, s increases up to 3 when all variables are 1. Therefore, perhaps the range is [2,3].But to confirm this, maybe use inequalities. For example, since each term ( frac{a + b}{1 + c} ), let's see if we can bound each term.Since a and b are at least 0.5, the numerator is at least 1. The denominator is at most 2 (since c is at most 1). So each term is at least 0.5. But since there are three terms, the total is at least 1.5. But this contradicts our earlier result. Wait, but maybe this approach is too rough.Alternatively, using the fact that variables are in [0.5,1], perhaps use the method of substitution. Let me set x = a, y = b, z = c, with x, y, z ∈ [0.5,1]. Then, s = (x + y)/(1 + z) + (y + z)/(1 + x) + (z + x)/(1 + y).Alternatively, maybe use the Cauchy-Schwarz inequality. But I'm not sure how to apply it here.Alternatively, note that when variables increase, the numerators in the terms increase and the denominators also increase. However, the effect might not be straightforward. For example, increasing a variable in the numerator of one term may increase another term's denominator. It's a bit tangled.Alternatively, let's consider the function s(a,b,c) and check if it's increasing or decreasing with respect to each variable. For example, partial derivative with respect to a:ds/da = [1/(1 + c)] + [ - (b + c)/(1 + a)^2 ] + [1/(1 + b)]Wait, this might be too complicated. Let me see.Alternatively, maybe assume that the extremum occurs at the endpoints, as in the cases when variables are 0.5 or 1. Since when we tried intermediate values, s was between 2 and 3. So, perhaps the minimal s is 2 and maximum is 3.Alternatively, let's use the method of substitution. Let me set variables as 0.5 + t, where t ∈ [0, 0.5]. Let me denote a = 0.5 + t1, b = 0.5 + t2, c = 0.5 + t3, where ti ∈ [0,0.5]. Then, s can be expressed in terms of ti. However, this might not lead to a straightforward solution.Alternatively, let's consider the sum s. Let's note that each term is of the form (a + b)/(1 + c). Let's see if we can relate this to a known inequality.Alternatively, note that when a, b, c are in [0.5,1], then 1 + c is in [1.5, 2], and a + b is in [1, 2]. Therefore, each term (a + b)/(1 + c) is in [1/2, 4/3]. But summing three such terms would give s ∈ [1.5, 4], but this is a rough estimate, and we already saw that s actually lies between 2 and 3.Alternatively, maybe use the Nesbitt's inequality. Wait, Nesbitt's inequality is (a/(b + c) + b/(a + c) + c/(a + b)) ≥ 3/2, but our expression is different.Alternatively, consider the following approach: For each term, since a, b ≥ 1/2 and c ≤1, then (a + b) ≥1, and (1 + c) ≤2, so (a + b)/(1 + c) ≥1/2. But again, summing three gives ≥1.5, which is not helpful.Alternatively, consider that when all variables are equal, s is 3*(2a)/(1 + a). Then, when a =0.5, s=3*(1)/(1.5)=2, and when a=1, s=3*(2)/2=3. So that's the same as before. If variables are equal, s ranges from 2 to3. But when variables are unequal, can s exceed these bounds? For example, when two variables are higher and one is lower, but we saw that in such cases, s is less than 3. Similarly, when two variables are lower and one is higher, s is higher than 2 but less than 3.Wait, actually, when variables are unequal, does s get larger or smaller? Let's see.Suppose two variables are higher and one is lower. Let's set a=1, b=1, c=0.5. Then s ≈ 2.833, which is less than 3. If two variables are lower and one is higher, like a=0.5, b=0.5, c=1, s=2.5, which is higher than 2 but less than 3. So, in both cases, s stays within [2,3]. Thus, it's possible that 2 and 3 are indeed the minimal and maximal values.Alternatively, let's use the method of Lagrange multipliers to find extrema. Let's assume that a, b, c are in (1/2,1). Then, set up the function:s = (a + b)/(1 + c) + (b + c)/(1 + a) + (c + a)/(1 + b)Compute partial derivatives with respect to a, b, c and set them equal to zero.First, ∂s/∂a = [1/(1 + c)] - (b + c)/(1 + a)^2 + [1/(1 + b)]Similarly, ∂s/∂b = [1/(1 + c)] + [1/(1 + a)] - (c + a)/(1 + b)^2∂s/∂c = [ - (a + b)/(1 + c)^2 ] + [1/(1 + a)] + [1/(1 + b)]Set each derivative to zero.But this system of equations might be difficult to solve. However, if we assume that a = b = c, then we can check if this is a critical point.Let’s set a = b = c = t. Then s = 3*(2t)/(1 + t) = 6t/(1 + t)Compute derivative with respect to t: 6*(1 + t) - 6t*(1) / (1 + t)^2 = 6 / (1 + t)^2 > 0. So s is increasing in t, which we saw earlier. So when t is 0.5, s=2, when t=1, s=3. Therefore, if the extremum occurs at a = b = c, then the maximum is at t=1 and minimum at t=0.5. However, we need to check if there are other critical points where variables are unequal.But given that when variables are unequal, s is between 2 and 3, and the derivative at equal variables is always positive, maybe the only extrema are at the endpoints. Therefore, the range of s is [2,3].Alternatively, to confirm, let's consider that each term (a + b)/(1 + c) is maximized when a and b are as large as possible and c as small as possible. But due to the symmetry, if we try to maximize one term by setting a=1, b=1, c=0.5, the other terms get smaller. Similarly, trying to minimize a term by setting a=0.5, b=0.5, c=1 makes other terms bigger. Thus, there is a balance, and the total sum s remains between 2 and 3.Therefore, after considering various cases and boundaries, it seems that the minimal value of s is 2 and the maximal value is 3. Therefore, the range of s is [2,3].Final AnswerThe range of ( s ) is (boxed{[2, 3]}).</think>Given ( a, b, c in left[ frac{1}{2}, 1 right] ), we need to find the range of the expression ( s = frac{a+b}{1+c} + frac{b+c}{1+a} + frac{c+a}{1+b} ).First, we consider the case where all variables are equal. If ( a = b = c = frac{1}{2} ), then each term in ( s ) becomes ( frac{1}{1.5} = frac{2}{3} ), leading to ( s = 3 times frac{2}{3} = 2 ). If ( a = b = c = 1 ), each term becomes ( frac{2}{2} = 1 ), leading to ( s = 3 times 1 = 3 ).Next, we check other configurations. For example, setting ( a = 1 ), ( b = 1 ), ( c = frac{1}{2} ) results in ( s approx 2.833 ). Setting ( a = 1 ), ( b = frac{1}{2} ), ( c = frac{1}{2} ) results in ( s = 2.5 ). These values are still between 2 and 3.We also consider the partial derivatives and critical points, but find that the function is symmetric and increasing when variables are equal. Testing various configurations where variables are at the endpoints (0.5 or 1) confirms that the minimum value of ( s ) is 2 and the maximum value is 3.Thus, the range of ( s ) is (boxed{[2, 3]}).]

🔑:3. Solution: Since a+b geqslant 1, then frac{a+b}{1+c} geqslant frac{a+b}{a+b+c}, thus frac{b+c}{1+a} geqslant frac{b+c}{a+b+c}, frac{c+a}{1+b} geqslant frac{c+a}{a+b+c}, adding them up, we get s geqslant 2, which is achieved when a=b=c=frac{1}{2}.Since a leqslant 1, then frac{a}{1+c} leqslant frac{a}{a+c}, frac{c}{1+a} leqslant frac{c}{a+c}, similarly,frac{b}{1+c} leqslant frac{b}{b+c}, frac{b}{1+a} leqslant frac{b}{a+b} ; frac{c}{1+b} leqslant frac{c}{b+c}, frac{a}{1+b} leqslantfrac{a}{a+b}, adding them up, we get s leqslant 3, which is achieved when a=b=c=1. therefore s in[2, 3].

🧠:[<think>Okay, let's try to tackle these problems one by one. Starting with problem (1): The solutions to the equation (x² + 2)/x = 5 + 2/5. Hmm. So according to the reading material, if I have an equation of the form x + (ab)/x = a + b, then the solutions are x₁ = a and x₂ = b. Let me see if I can rewrite the given equation in that form.First, let's simplify the right-hand side. 5 + 2/5 is equal to... 5 is 25/5, so 25/5 + 2/5 = 27/5. So the equation is (x² + 2)/x = 27/5. Let me rearrange the left side. Dividing by x, so that's x²/x + 2/x = x + 2/x. So the equation becomes x + 2/x = 27/5. Comparing this to the standard form x + (ab)/x = a + b from the reading material. Here, (ab) would be 2, and a + b would be 27/5. So according to the material, the solutions should be x₁ = a and x₂ = b. So if I can find a and b such that a + b = 27/5 and ab = 2. Then the solutions are a and b. Wait, but in the standard problem, the equation is x + (ab)/x = a + b, so here our equation is x + 2/x = 27/5. So in this case, ab = 2 and a + b = 27/5. So the solutions should be x = a and x = b where a and b satisfy those two conditions. Therefore, the solutions are 5 and 2/5? Wait, let me check. If a + b = 27/5 and ab = 2, then solving these two equations. Let's let a and b be the roots of the quadratic equation t² - (a + b)t + ab = 0, which would be t² - (27/5)t + 2 = 0. Multiplying through by 5 to eliminate the fraction: 5t² - 27t + 10 = 0. Let's factor this. Looking for two numbers that multiply to 5*10=50 and add up to -27. Hmm, 50 can be split into -25 and -2, which multiply to 50 and add to -27. So 5t² -25t -2t +10 =0. Then, group as (5t² -25t) + (-2t +10) =0. Factor out 5t from the first group: 5t(t -5) -2(t -5)=0. Then (5t -2)(t -5)=0. So t = 5 or t = 2/5. Therefore, the solutions are x₁=5 and x₂=2/5. So that should be the answer for part (1). Let me just verify by plugging back into the original equation. For x=5: (25 + 2)/5 = 27/5, which is 5.4. The right side is 5 + 2/5 = 5.4, so that checks out. For x=2/5: ( (4/25) + 2 ) divided by (2/5). Let's compute numerator: 4/25 + 2 = 4/25 + 50/25 = 54/25. Divided by 2/5 is 54/25 * 5/2 = 54/10 = 27/5, which is also 5.4, matching the right-hand side. So that's correct. Therefore, x₁=5 and x₂=2/5.Moving on to problem (2): If the solutions to x + 3/x =7 are x₁=a and x₂=b, find a² + b². From the reading material, we know that if the equation is x + (ab)/x = a + b, then solutions are x=a and x=b. Comparing this to the given equation x + 3/x =7. So here, ab=3 and a + b=7. Therefore, the solutions x=a and x=b are such that a + b=7 and ab=3. The question is asking for a² + b². Recall that a² + b² = (a + b)² - 2ab. Plugging in the known values: (7)² - 2*(3) = 49 -6=43. Therefore, a² + b²=43. Let me verify this. Suppose a and b are roots of the equation x +3/x=7. Multiply both sides by x: x² +3 =7x => x² -7x +3=0. The roots of this quadratic are a and b. The sum of roots is 7, product is 3. Then indeed, a² + b²=(a + b)^2 -2ab=49 -6=43. So that checks out. Problem (3): If the solutions to the equation 6/(x -1) =k -x are x₁=t +1 and x₂= t² +2, find the value of k² -4k +4t³. Hmm. This seems a bit more involved. Let's try to approach this step by step.First, the equation is 6/(x -1) =k -x. Let me rewrite this equation. Multiply both sides by (x -1) to eliminate the denominator: 6 = (k -x)(x -1). Expand the right-hand side: (k -x)(x -1) =k*x -k -x² +x. So 6 = -x² + (k +1)x -k. Bring all terms to one side: -x² + (k +1)x -k -6=0. Multiply through by -1 to make the quadratic coefficient positive: x² - (k +1)x +k +6=0. So the quadratic equation is x² - (k +1)x + (k +6)=0. The solutions to this equation are given as x₁ = t +1 and x₂= t² +2. Therefore, according to Vieta's formulas, the sum of the roots x₁ + x₂ = (k +1) and the product x₁x₂ = k +6. Let's compute sum and product of the roots.Sum: (t +1) + (t² +2) = t² + t +3. So according to Vieta, t² + t +3 = k +1. Therefore, t² + t +3 =k +1 => k = t² + t +2.Product: (t +1)(t² +2) = t*(t² +2) +1*(t² +2) = t³ +2t + t² +2 = t³ + t² +2t +2. According to Vieta, this product equals k +6. Therefore, t³ + t² +2t +2 = k +6. But we already have k expressed in terms of t: k = t² + t +2. So substitute that into the equation: t³ + t² +2t +2 = (t² + t +2) +6. Simplify the right-hand side: t² + t +2 +6= t² + t +8. Therefore, the equation becomes t³ + t² +2t +2 = t² + t +8. Subtract t² + t +8 from both sides: t³ + t² +2t +2 -t² -t -8= t³ + t -6=0. So t³ + t -6=0. Let's solve this cubic equation for t. Looking for rational roots, possible candidates are ±1, ±2, ±3, ±6. Let's test t=1: 1 +1 -6= -4 ≠0. t=2: 8 +2 -6=4≠0. t=3:27 +3 -6=24≠0. t= -1: -1 -1 -6=-8≠0. t= -2: -8 -2 -6=-16≠0. t=6:216 +6 -6=216≠0. Hmm, none of these work. Maybe we need to factor this cubic. Alternatively, perhaps made a miscalculation. Let me check the steps again.Original equation: 6/(x -1)=k -x. Multiply by (x -1): 6=(k -x)(x -1). Expand RHS: kx -k -x² +x. So 6= -x² + (k +1)x -k. Bring all terms to left: -x² + (k +1)x -k -6=0. Multiply by -1: x² - (k +1)x +k +6=0. Correct.Sum of roots: x₁ +x₂= (t +1) + (t² +2)=t² +t +3. Vieta's sum is (k +1). So k = t² + t +2. Correct.Product: (t +1)(t² +2)=t³ +2t + t² +2= t³ +t² +2t +2. Vieta's product is k +6. So t³ +t² +2t +2 =k +6. Since k = t² + t +2, substitute: t³ +t² +2t +2 = t² +t +2 +6. RHS becomes t² +t +8. So t³ +t² +2t +2 -t² -t -8= t³ + t -6=0. Correct. So t³ + t -6=0.Hmm, no rational roots. Maybe we need to solve this cubic. Alternatively, perhaps made a mistake in interpreting the problem. Let me check once more.Wait, maybe the equation after substitution is t³ + t -6=0. So need to find real solution for t. Let's try t=2: 8 +2 -6=4≠0. Wait, t=1.5: (3.375) +1.5 -6= -1.125. t=1.8: (5.832) +1.8 -6=1.632. t=1.7: (4.913) +1.7 -6=0.613. t=1.6: (4.096)+1.6 -6= -0.304. So between 1.6 and 1.7, there's a root. But this seems messy. Wait, but maybe there's a different approach.Alternatively, since the problem asks for k² -4k +4t³. Let's see if we can express this in terms of t. We have k = t² + t +2. So let's compute k² -4k +4t³.First, compute k²: (t² +t +2)^2 = t^4 + 2t³ + (t² +4t²) +4t +4. Wait, expanding (t² + t +2)^2:= t²(t² + t +2) + t(t² + t +2) +2(t² + t +2)= t^4 + t³ +2t² + t³ +t² +2t +2t² +2t +4Combine like terms:t^4 + (t³ + t³) + (2t² +t² +2t²) + (2t +2t) +4= t^4 + 2t³ +5t² +4t +4.Then, compute 4k: 4*(t² +t +2)=4t² +4t +8.So, k² -4k = t^4 +2t³ +5t² +4t +4 - (4t² +4t +8) = t^4 +2t³ + (5t² -4t²) + (4t -4t) + (4 -8) = t^4 +2t³ +t² -4.Then, the expression k² -4k +4t³ is t^4 +2t³ +t² -4 +4t³ = t^4 +6t³ +t² -4.But we also have the equation from before: t³ +t -6=0. Let's write t³=6 -t. So perhaps substitute t³=6 -t into the expression t^4 +6t³ +t² -4.First, compute t^4. t^4 = t*t³ = t*(6 -t)=6t -t².So substitute into t^4 +6t³ +t² -4:(6t -t²) +6*(6 -t) +t² -4=6t -t² +36 -6t +t² -4Simplify:6t -6t cancels. -t² +t² cancels. 36 -4=32. Therefore, the entire expression simplifies to 32. Wow, that's neat. So despite the cubic equation being difficult to solve, by substituting t³=6 -t, we can compute the expression k² -4k +4t³ as 32. Therefore, the answer is 32. Let me verify this step-by-step.Given t³ =6 -t.Expression: k² -4k +4t³. We found k = t² + t +2. Then k² -4k = (t² + t +2)^2 -4(t² +t +2). Then added 4t³. But when expanding k² -4k, we had t^4 +2t³ +5t² +4t +4 -4t² -4t -8 = t^4 +2t³ +t² -4. Then adding 4t³: t^4 +6t³ +t² -4. Now substitute t^4 = t*(t³) = t*(6 -t) =6t -t². So replacing t^4 with 6t -t²:6t -t² +6t³ +t² -4 =6t +6t³ -4. But since t³=6 -t, replace t³:6t +6*(6 -t) -4 =6t +36 -6t -4=32. Yes, that works out. So regardless of the value of t, the expression k² -4k +4t³ equals 32. Therefore, the answer is 32.Final Answer(1) The solutions are boxed{5} and boxed{dfrac{2}{5}}; (2) The value of a^{2}+b^{2} is boxed{43}; (3) The value of k^{2}-4k+4t^{3} is boxed{32}.</think>(1) We start with the equation (frac{x^2 + 2}{x} = 5 + frac{2}{5}). Simplifying the right-hand side, we get (5 + frac{2}{5} = frac{27}{5}). Rewriting the equation, we have (x + frac{2}{x} = frac{27}{5}). Comparing this to the standard form (x + frac{ab}{x} = a + b), we identify (ab = 2) and (a + b = frac{27}{5}). Solving the quadratic equation (t^2 - frac{27}{5}t + 2 = 0), we find the roots (t = 5) and (t = frac{2}{5}). Thus, the solutions are (x_1 = 5) and (x_2 = frac{2}{5}).(2) Given the equation (x + frac{3}{x} = 7), we identify (a + b = 7) and (ab = 3). Using the identity (a^2 + b^2 = (a + b)^2 - 2ab), we substitute the known values: (a^2 + b^2 = 7^2 - 2 cdot 3 = 49 - 6 = 43).(3) Starting with the equation (frac{6}{x - 1} = k - x), we multiply both sides by (x - 1) to get (6 = (k - x)(x - 1)). Expanding and rearranging, we obtain the quadratic equation (x^2 - (k + 1)x + (k + 6) = 0). The solutions (x_1 = t + 1) and (x_2 = t^2 + 2) give us the sum and product of the roots: (t^2 + t + 3 = k + 1) and (t^3 + t^2 + 2t + 2 = k + 6). Solving for (k), we find (k = t^2 + t + 2). Substituting (k) into the product equation, we derive (t^3 + t - 6 = 0). Using (t^3 = 6 - t), we compute (k^2 - 4k + 4t^3) by substituting (k = t^2 + t + 2) and simplifying, resulting in the final value 32. Final Answer(1) The solutions are (boxed{5}) and (boxed{dfrac{2}{5}}); (2) The value of (a^{2}+b^{2}) is (boxed{43}); (3) The value of (k^{2}-4k+4t^{3}) is (boxed{32}).,<think>Okay, let's tackle these three problems one by one. Starting with problem (1). The equation given is (x² + 2)/x = 5 + 2/5. Hmm, first, I need to simplify both sides of this equation. Let me rewrite the right side: 5 + 2/5 is equal to 25/5 + 2/5, which is 27/5. So the equation becomes (x² + 2)/x = 27/5. Now, according to the reading material, if we have an equation of the form x + ab/x = a + b, then the solutions are x1 = a and x2 = b. Let me check if I can manipulate the given equation into that form. Let's see. Starting with (x² + 2)/x, if I split the numerator, that's x²/x + 2/x = x + 2/x. So the left side simplifies to x + 2/x. The right side is 27/5. Therefore, the equation is x + 2/x = 27/5. Comparing this to the standard form given in the reading material, which is x + (ab)/x = a + b, here ab would be 2, and a + b is 27/5. So, according to the reading, the solutions are x1 = a and x2 = b, where a and b are numbers such that a + b = 27/5 and ab = 2. So, in this case, a and b are the solutions to the system of equations:a + b = 27/5ab = 2But wait, according to the reading material, the solutions to the equation x + ab/x = a + b are x = a and x = b. So, does that mean that in our equation, since it's x + 2/x = 27/5, the solutions are x1 = a and x2 = b, where a and b satisfy a + b = 27/5 and ab = 2? Therefore, the solutions are just the roots of the quadratic equation x² - (a + b)x + ab = 0, which would be x² - (27/5)x + 2 = 0. Solving this quadratic should give the solutions. Let me check that.Alternatively, maybe we can cross-multiply to solve the equation (x² + 2)/x = 27/5. Multiplying both sides by x (assuming x ≠ 0), we get x² + 2 = (27/5)x. Then, bringing all terms to one side: x² - (27/5)x + 2 = 0. Multiply all terms by 5 to eliminate the fraction: 5x² - 27x + 10 = 0. Now, solve this quadratic equation. Let's use the quadratic formula. The discriminant D is (-27)^2 - 4*5*10 = 729 - 200 = 529. Square root of 529 is 23. Therefore, the solutions are [27 ± 23]/(2*5) = (27 + 23)/10 = 50/10 = 5, and (27 - 23)/10 = 4/10 = 2/5. So the solutions are x1 = 5 and x2 = 2/5. But according to the reading material, if we have x + ab/x = a + b, the solutions are x = a and x = b. So here, ab is 2 and a + b is 27/5. So the solutions should be the a and b such that their sum is 27/5 and product is 2. Indeed, if a = 5 and b = 2/5, then a + b = 5 + 0.4 = 5.4 = 27/5, and a*b = 5*(2/5) = 2, which matches. Therefore, the solutions are 5 and 2/5. So problem (1) answers are x1 = 5 and x2 = 2/5.Moving on to problem (2). If the solutions to the equation x + 3/x = 7 are x1 = a and x2 = b, find a² + b². Okay, so here the equation is x + 3/x = 7. According to the reading material, solutions are x1 = a and x2 = b where a + b = 7 and ab = 3. Wait, no. Wait, in the reading material, the equation is x + (ab)/x = a + b, so here, comparing, we have x + 3/x = 7. Therefore, in this case, a + b = 7 and ab = 3. Therefore, the solutions are x = a and x = b where a and b are roots of the quadratic equation x² - 7x + 3 = 0. So a and b are the roots, so to find a² + b², we can use the identity a² + b² = (a + b)² - 2ab. Since a + b = 7 and ab = 3, then a² + b² = 7² - 2*3 = 49 - 6 = 43. Therefore, the answer is 43.But let me verify this by actually solving the equation. The equation x + 3/x = 7. Multiply both sides by x: x² + 3 = 7x. Rearranged: x² -7x +3 =0. The solutions are [7 ± sqrt(49 -12)]/2 = [7 ± sqrt(37)]/2. Let's denote a = [7 + sqrt(37)]/2 and b = [7 - sqrt(37)]/2. Then a² + b². Let's compute it:First, (a + b) = 7, ab = 3. So a² + b² = (a + b)^2 - 2ab = 49 - 6 = 43. Yep, same result. So problem (2) answer is 43.Now problem (3). If the solutions to the equation 6/(x -1) = k - x are x1 = t +1 and x2 = t² + 2, find the value of k² -4k +4t³. Hmm. This seems more complicated. Let's see.First, the equation is 6/(x - 1) = k - x. Let's try to solve this equation for x and see how the roots relate to t and the given expressions. The solutions are given as x1 = t +1 and x2 = t² + 2. So perhaps we need to express k in terms of t, and then compute the expression k² -4k +4t³.First, let's solve the equation 6/(x -1) = k - x. Let's multiply both sides by (x -1) to eliminate the denominator: 6 = (k - x)(x -1). Expand the right-hand side: (k -x)(x -1) = kx -k -x² +x. Therefore, the equation becomes 6 = kx -k -x² +x. Let's rearrange all terms to one side: -x² + (k +1)x -k -6 = 0. Multiply both sides by -1 to make it standard: x² - (k +1)x +k +6 =0.So the quadratic equation is x² - (k +1)x + (k +6) =0. The roots of this equation are given as x1 = t +1 and x2 = t² +2. So using Vieta's formula, the sum of the roots x1 + x2 = (k +1) and the product x1*x2 = k +6.So let's compute the sum and product of the roots:Sum: (t +1) + (t² +2) = t +1 + t² +2 = t² + t +3. This should equal k +1.Product: (t +1)(t² +2) = t*(t² +2) +1*(t² +2) = t³ +2t + t² +2 = t³ + t² +2t +2. This should equal k +6.So from the sum: t² + t +3 = k +1 ⇒ k = t² + t +3 -1 = t² + t +2.From the product: t³ + t² +2t +2 = k +6. Substitute the value of k from above:t³ + t² +2t +2 = (t² + t +2) +6 ⇒ t³ + t² +2t +2 = t² + t +8.Subtract t² + t +8 from both sides:t³ + t² +2t +2 - t² -t -8 = t³ + (t² - t²) + (2t -t) + (2 -8) = t³ + t -6 =0.Therefore, t³ + t -6 =0. Let's solve this equation for t. Maybe there's a real root we can find. Let's test t=1: 1 +1 -6= -4≠0. t=2: 8 +2 -6=4≠0. t=3:27 +3 -6=24≠0. t=√6? Probably not. Wait, maybe there's a rational root. Using rational root theorem, possible roots are ±1, ±2, ±3, ±6. Let's try t=2: 8 +2 -6=4≠0. t=1: -4, t=3:24. t= -1: -1 -1 -6=-8. t= -2: -8 -2 -6=-16. Hmm, none of these. So perhaps this cubic doesn't factor nicely. Wait, but the problem doesn't ask for t, but rather the expression k² -4k +4t³. Let's see. We have k = t² + t +2, and we need to compute k² -4k +4t³. Let's try to express everything in terms of t.First, compute k²: (t² + t +2)^2. Let's expand that:= t^4 + 2t^3 + (t^2 +4t +4) + 4t² +4t +4. Wait, no. Wait, (a + b + c)^2 = a² + b² + c² + 2ab + 2ac + 2bc. So here, a = t², b = t, c = 2. So:= (t²)^2 + t^2 + 2^2 + 2*(t²*t) + 2*(t²*2) + 2*(t*2)= t^4 + t^2 +4 + 2t^3 +4t² +4tCombine like terms:= t^4 +2t^3 + (t^2 +4t^2) +4t +4= t^4 +2t^3 +5t² +4t +4Then subtract 4k:-4k = -4*(t² + t +2) = -4t² -4t -8Add 4t³:+4t³So putting it all together:k² -4k +4t³ = [t^4 +2t^3 +5t² +4t +4] + [-4t² -4t -8] +4t³Combine like terms:t^4 + (2t³ +4t³) + (5t² -4t²) + (4t -4t) + (4 -8)= t^4 +6t³ +t² +0t -4So k² -4k +4t³ = t^4 +6t³ +t² -4But we also have from the product equation: t³ + t -6 =0 ⇒ t³ = -t +6. Let's see if we can substitute t³ in the above expression. Let's compute t^4 +6t³ +t² -4.First, t^4 = t*t³. Since t³ = -t +6, so t^4 = t*(-t +6) = -t² +6t.Similarly, 6t³ =6*(-t +6) = -6t +36.So substitute back into the expression:t^4 +6t³ +t² -4 = (-t² +6t) + (-6t +36) +t² -4Simplify:(-t² + t²) + (6t -6t) + (36 -4) = 0 +0 +32 =32.Therefore, k² -4k +4t³ =32.Wait, that's neat. So by using the relation t³ = -t +6, we can reduce the expression t^4 +6t³ +t² -4 to 32. Therefore, the answer is 32. Let me check this again.First, from t³ = -t +6.Compute t^4: t*t³ = t*(-t +6) = -t² +6t.Compute 6t³:6*(-t +6) = -6t +36.So, t^4 +6t³ +t² -4 becomes (-t² +6t) + (-6t +36) + t² -4.Combine like terms:(-t² + t²) =06t -6t=036 -4=32Therefore, yes, the entire expression simplifies to 32. So k² -4k +4t³=32.Therefore, the answer is 32.Final Answer(1) The solutions are boxed{5} and boxed{dfrac{2}{5}}; (2) The value of a^2 + b^2 is boxed{43}; (3) The value of k^2 -4k +4t^3 is boxed{32}.</think>(1) The given equation is (frac{x^2 + 2}{x} = 5 + frac{2}{5}). Simplifying the right-hand side, we get (frac{27}{5}). Rewriting the equation, we have (x + frac{2}{x} = frac{27}{5}). By comparing this with the form (x + frac{ab}{x} = a + b), we identify (a + b = frac{27}{5}) and (ab = 2). Solving the quadratic equation (x^2 - frac{27}{5}x + 2 = 0) using the quadratic formula, we find the solutions (x = 5) and (x = frac{2}{5}).(2) The given equation is (x + frac{3}{x} = 7). Here, (a + b = 7) and (ab = 3). Using the identity (a^2 + b^2 = (a + b)^2 - 2ab), we substitute the values to get (a^2 + b^2 = 7^2 - 2 cdot 3 = 49 - 6 = 43).(3) The given equation is (frac{6}{x - 1} = k - x). Multiplying both sides by (x - 1) and rearranging, we get the quadratic equation (x^2 - (k + 1)x + k + 6 = 0). The roots are (x_1 = t + 1) and (x_2 = t^2 + 2). Using Vieta's formulas, we find (k = t^2 + t + 2) and substitute into the expression (k^2 - 4k + 4t^3). By solving the cubic equation (t^3 + t - 6 = 0) and substituting (t^3 = -t + 6), we simplify the expression to find the final value (32). Final Answer(1) The solutions are (boxed{5}) and (boxed{dfrac{2}{5}}); (2) The value of (a^2 + b^2) is (boxed{43}); (3) The value of (k^2 - 4k + 4t^3) is (boxed{32}).]

🔑: Step-by-Step Solution# (1) Understanding and ApplicationGiven the equation frac{x^2 + 2}{x} = 5 + frac{2}{5}, we can rewrite it as x + frac{2}{x} = 5 + frac{2}{5} by understanding the form x + frac{ab}{x} = a + b.- The solutions to the equation x + frac{2}{x} = 5 + frac{2}{5} are found by setting x = 5 or x = frac{2}{5}, based on the given form.Therefore, the solutions are x_1 = 5 and x_2 = frac{2}{5}, which can be encapsulated as boxed{x_1 = 5, x_2 = frac{2}{5}}.# (2) Knowledge TransferGiven the equation x + frac{3}{x} = 7, we identify that a + b = 7 and ab = 3.- To find a^2 + b^2, we use the identity (a + b)^2 = a^2 + b^2 + 2ab.- Thus, a^2 + b^2 = (a + b)^2 - 2ab = 7^2 - 2 times 3 = 49 - 6 = 43.The value of a^2 + b^2 is boxed{43}.# (3) Extension and EnhancementGiven the equation frac{6}{x - 1} = k - x, we can transform it to x - 1 + frac{6}{x - 1} = k - 1.- Letting y = x - 1, the equation becomes y + frac{6}{y} = k - 1.- From this, we know y_1 cdot y_2 = 6 and y_1 + y_2 = k - 1.- Given x_1 = t + 1 and x_2 = t^2 + 2, we find y_1 = t and y_2 = t^2 + 1.- This leads to t(t^2 + 1) = 6 and t + t^2 + 1 = k - 1.- Therefore, k = t + t^2 + 2 and t^3 + t = 6.To find k^2 - 4k + 4t^3, we calculate:- k^2 - 4k + 4t^3 = (t + t^2 + 2)^2 - 4(t + t^2 + 2) + 4t^3- = t^4 + 2t^3 + t^2 + 4t^3 + 4t^2 + 4 - 4t - 4t^2 - 8 + 4t^3- = t^4 + 6t^3 + 5t^2 - 4t - 4- Since t^3 + t = 6, we substitute to get 6t + 6t^3 - 4- = 6(6) - 4 = 36 - 4 = 32.The value of k^2 - 4k + 4t^3 is boxed{32}.

🧠:[<think>Okay, let's tackle this problem step by step. First, I need to understand what the question is asking. We have two sets, M and N, defined by certain conditions, and we need to find the intersection of the complement of M with N. The universal set is all real numbers, U = ℝ. The answer choices are given, so after solving, I can check which one matches.Starting with set M: M is defined as {x | y = ln(1 - x)}. Hmm, the notation here is a bit confusing. Usually, when we define a set like {x | condition}, the condition should be a statement about x. But here it says y = ln(1 - x). Maybe this is a typo or misinterpretation? Wait, perhaps they mean that M is the domain of the function y = ln(1 - x). Because the natural logarithm function ln(z) is only defined when z > 0. Therefore, for ln(1 - x) to be defined, the argument 1 - x must be greater than 0. So 1 - x > 0 implies x < 1. Therefore, the domain of the function y = ln(1 - x) is all real numbers x where x < 1. So set M is {x | x < 1}.Okay, that makes sense. So M is all real numbers less than 1. Therefore, the complement of M in the universal set U (which is all real numbers) would be all real numbers not in M. Since M is x < 1, the complement C_U M is x ≥ 1. Got it. So complement of M is [1, ∞).Now moving on to set N: N is defined as {x | 2^{x(x - 2)} < 1}. Let's parse this inequality. 2 raised to some exponent is less than 1. Since 2 is a positive number greater than 1, 2^a < 1 if and only if the exponent a is negative. Because when you raise 2 to a negative power, you get 1/(2^|a|), which is less than 1. So the inequality 2^{x(x - 2)} < 1 is equivalent to x(x - 2) < 0.So solving x(x - 2) < 0. Let's find the critical points where the expression equals zero: x = 0 and x = 2. These divide the real number line into intervals. Let's test each interval:1. When x < 0: Let's pick x = -1. Then (-1)(-1 - 2) = (-1)(-3) = 3 > 0. So positive in this interval.2. When 0 < x < 2: Let's pick x = 1. Then (1)(1 - 2) = (1)(-1) = -1 < 0. So negative here.3. When x > 2: Let's pick x = 3. Then (3)(3 - 2) = (3)(1) = 3 > 0. Positive here.Since we want x(x - 2) < 0, the solution is the interval where the expression is negative, which is (0, 2). Therefore, set N is {x | 0 < x < 2}.Now, we need to find the intersection of the complement of M with N. The complement of M is [1, ∞), and N is (0, 2). The intersection of these two sets is the overlap between [1, ∞) and (0, 2). So [1, ∞) ∩ (0, 2) would be [1, 2). Because from 0 to 2, the overlap with [1, ∞) starts at 1 and goes up to 2.Therefore, (C_U M) ∩ N is {x | 1 ≤ x < 2}. Looking at the answer choices, that's option B.But let me double-check each step to make sure I didn't make a mistake.First, for set M: Since ln(1 - x) is defined when 1 - x > 0, so x < 1. That's correct. So M is (-∞, 1), complement is [1, ∞). Okay.For set N: The inequality 2^{x(x - 2)} < 1. Since 2^a < 1 implies a < 0, so x(x - 2) < 0. Solving the quadratic inequality, the critical points at 0 and 2, test intervals. Correct, the solution is (0, 2). So N is (0, 2). Right.Then the intersection: [1, ∞) intersect (0, 2) is [1, 2). So yes, that's {x | 1 ≤ x < 2}, which is option B. The answer should be B.But let me check the answer choices again to confirm:A: {x | x ≥ 1}B: {x | 1 ≤ x < 2}C: {x | 0 ≤ x < 1}D: {x | 0 < x ≤1}Yes, B matches our result.Wait, but just to ensure there's no trick here. For the set M, the problem says "M = {x | y = ln(1 - x)}". If we interpret this as the set of all x such that there exists a y where y = ln(1 - x), then that's equivalent to the domain of the function, which is x < 1. So that's correct.Similarly, for set N: {x | 2^{x(x - 2)} < 1}. Since 2^{x(x - 2)} is defined for all real x because exponentials are defined for all real exponents. So the inequality is straightforward, and solving it as we did gives x between 0 and 2. Correct.Therefore, the intersection is indeed [1, 2), which is option B.Final Answerboxed{B}</think>Given the universal set ( U = mathbb{R} ), we need to find ((C_{U}M) cap N).1. Set ( M ): - ( M ) is defined as ({x mid y = ln(1 - x)}). - The natural logarithm function (ln(z)) is defined for ( z > 0 ). Therefore, ( 1 - x > 0 ) implies ( x < 1 ). - Hence, ( M = {x mid x < 1} ). - The complement of ( M ) in ( U ) is ( C_{U}M = {x mid x geq 1} ).2. Set ( N ): - ( N ) is defined as ({x mid 2^{x(x-2)} < 1}). - The inequality ( 2^{x(x-2)} < 1 ) holds if and only if the exponent ( x(x - 2) ) is negative. - Solving ( x(x - 2) < 0 ), we find the critical points ( x = 0 ) and ( x = 2 ). Testing intervals, the solution is ( 0 < x < 2 ). - Hence, ( N = {x mid 0 < x < 2} ).3. Intersection ((C_{U}M) cap N): - ( C_{U}M = {x mid x geq 1} ) which is ([1, infty)). - ( N = {x mid 0 < x < 2} ) which is ((0, 2)). - The intersection of ([1, infty)) and ((0, 2)) is ([1, 2)).Thus, the answer is (boxed{B}).,<think>Okay, let me try to figure out this problem step by step. So, the question is about finding the intersection of the complement of set M and set N, given that the universal set is all real numbers. The answer choices are A to D, and I need to pick the correct one. Let me start by understanding what each set M and N represents.First, set M is defined as {x | y = ln(1 - x)}. Hmm, wait, that notation might be a bit confusing. When they say {x | y = ln(1 - x)}, does that mean the set of all x values for which the expression ln(1 - x) is defined? Because y is a function of x here, right? So, for y to be a real number, the argument of the natural logarithm must be positive. That means 1 - x > 0. Solving that inequality gives x < 1. Therefore, set M is all real numbers x such that x < 1. So M = (-∞, 1).Okay, that makes sense. So the complement of M in the universal set U (which is all real numbers) would be all the x's not in M. Since M is x < 1, the complement C_U M would be x ≥ 1. So C_U M = [1, ∞). Got that.Now let's look at set N. It's defined as {x | 2^{x(x - 2)} < 1}. I need to solve this inequality. Remember that 2^a < 1 is equivalent to a < 0 because 2 raised to any real number is positive, and it's less than 1 when the exponent is negative. So, 2^{x(x - 2)} < 1 implies x(x - 2) < 0.Let me solve the inequality x(x - 2) < 0. To find where the product is negative, I can look at the critical points where each factor is zero. The critical points are x = 0 and x = 2. The intervals to test are (-∞, 0), (0, 2), and (2, ∞). For x < 0: Let's pick x = -1. Then (-1)(-1 - 2) = (-1)(-3) = 3 > 0. So positive.For 0 < x < 2: Let's pick x = 1. Then (1)(1 - 2) = (1)(-1) = -1 < 0. So negative.For x > 2: Let's pick x = 3. Then (3)(3 - 2) = (3)(1) = 3 > 0. So positive.Since the inequality is looking for where the product is less than 0, the solution is the interval where the product is negative, which is (0, 2). Therefore, set N is (0, 2). So N = {x | 0 < x < 2}.Now, we need to find the intersection of C_U M and N. C_U M is [1, ∞) and N is (0, 2). The intersection of these two sets would be the overlap between them. Let's visualize this:C_U M: [1, ∞)N: (0, 2)The overlap is from 1 up to 2. But since N is open at 2, the upper bound is 2 not included. So the intersection is [1, 2). In set notation, this is {x | 1 ≤ x < 2}.Looking at the answer choices:A: {x | x ≥ 1} → This includes all numbers from 1 to infinity, which isn't correct because N only goes up to 2.B: {x | 1 ≤ x < 2} → This matches our result.C: {x | 0 ≤ x < 1} → This is the intersection of N with M's complement? Wait, no. The complement of M is [1, ∞), so intersecting with N (0,2) would be [1,2). So C is 0 to 1, which is not part of the complement of M.D: {x | 0 < x ≤1} → Similar issue as C. This is part of M's complement only at x=1, but mostly in M.Therefore, the correct answer should be B.Wait, let me double-check to make sure I didn't make any mistakes in solving the inequalities.For set M: ln(1 - x) is defined when 1 - x > 0 ⇒ x < 1. So M is (-∞, 1). Its complement is [1, ∞). That seems right.For set N: 2^{x(x - 2)} < 1. Since 2^a < 1 ⇒ a < 0. So x(x - 2) < 0. The critical points at 0 and 2. Testing intervals: negative between 0 and 2. So N is (0, 2). Correct.Intersection of [1, ∞) and (0, 2) is [1, 2). Which is exactly answer B. So yes, B is the right choice.Just to make sure, let's check if any steps could have errors. When solving for the complement of M, since M is all x < 1, complement is x ≥ 1. When solving for N, the inequality 2^{x(x - 2)} < 1 leads to x(x - 2) < 0, which is correct because 2^k < 1 iff k < 0. The quadratic inequality solution is (0, 2), which is correct. Intersection between [1, ∞) and (0, 2) is indeed [1, 2). So answer B is correct.Final Answerboxed{B}</think>Given the universal set ( U = mathbb{R} ), we need to find ((C_U M) cap N).1. Set ( M ): - ( M = { x mid y = ln(1 - x) } ) - The natural logarithm function ( ln(1 - x) ) is defined when ( 1 - x > 0 ), which implies ( x < 1 ). - Therefore, ( M = (-infty, 1) ).2. Complement of ( M ): - The complement of ( M ) in ( U ) is ( C_U M = [1, infty) ).3. Set ( N ): - ( N = { x mid 2^{x(x-2)} < 1 } ) - The inequality ( 2^{x(x-2)} < 1 ) holds when the exponent ( x(x - 2) < 0 ). - Solving ( x(x - 2) < 0 ), we find the critical points ( x = 0 ) and ( x = 2 ). Testing intervals, the solution is ( (0, 2) ). - Therefore, ( N = (0, 2) ).4. Intersection of ( C_U M ) and ( N ): - ( C_U M = [1, infty) ) - ( N = (0, 2) ) - The intersection is ( [1, infty) cap (0, 2) = [1, 2) ).Thus, the correct answer is (boxed{B}).]

🔑:M={x|y=ln(1-x)}=(-infty,1), N={x|2^{x(x-2)}<1}=(0,2),Since the universal set U=mathbb{R}, then C_{U}M=[1,+infty)(C_{U}M) cap N=[1,+infty) cap (0,2)=[1,2)Therefore, the correct option is boxed{text{B}}.

🧠:[<think>Okay, so the problem is about Gray having t dollars, and this money is made up of 1 and 2 coins. The key point here is that she has the same number of 1 and 2 coins. The question is asking how many 1 coins she has. Hmm, let's see. Let me try to break this down step by step.First, let me note down the given information. Gray has t dollars. The total amount comes from 1 and 2 coins. The number of 1 coins is equal to the number of 2 coins. We need to find the number of 1 coins. Let's denote the number of 1 coins as x. Since the number of 2 coins is the same, that would also be x. Now, each 1 coin is worth 1, so x of them would contribute x dollars. Each 2 coin is worth 2, so x of those would contribute 2x dollars. Therefore, the total amount t should be the sum of these two contributions. So, the total amount is x (from the 1 coins) plus 2x (from the 2 coins). That would be x + 2x = 3x. So, according to this, t = 3x. Therefore, solving for x, which is the number of 1 coins, we get x = t / 3. Wait, but is this all? Let me check if I missed anything. The problem says Gray has t dollars consisting of 1 and 2 coins. The number of each type of coin is the same. So, if there are x 1 coins, there are x 2 coins. Each 1 coin is 1, so total from 1 coins is x. Each 2 coin is 2, so total from 2 coins is 2x. Summing them gives 3x = t. Therefore, x = t / 3. But hold on, the problem mentions "coins," so x must be an integer, right? Because you can't have a fraction of a coin. So, this would imply that t must be divisible by 3. If t is not divisible by 3, then such a situation isn't possible. However, the problem doesn't specify any constraints on t, so maybe we just proceed with the algebraic solution, assuming that t is a multiple of 3. But the problem says "TNYWR" for t. Wait, the original problem starts with "Let t be TNYWR." Hmm, "TNYWR" – maybe that's an abbreviation? Let me think. In some contexts, "TNYCup" is "The New York Cup," but "TNYWR" doesn't ring a bell. Wait, perhaps it's a typo? The original problem might have been in another language, and maybe "TNYWR" is a mistranslation or misprint. Alternatively, maybe it's an acronym for something. Let me check the original problem again.The problem states: "Let t be TNYWR. Gray has t dollars consisting of 1 and 2 coins. If she has the same number of 1 and 2 coins, how many 1 coins does she have?" Wait, maybe "TNYWR" is supposed to be a placeholder, like "a certain number" or "a specific value"? But in math problems, usually, variables are defined. So maybe "TNYWR" is supposed to be a typo or an abbreviation for something. Alternatively, maybe it's "TNYCup" or "TNYWRK" (The New York Work?), but I'm not sure. Alternatively, maybe it's a misinput, and the original problem had a specific value for t, but here it's replaced with "TNYWR." Since the problem is presented as is, and the user is asking me to solve it, perhaps I should proceed under the assumption that "TNYWR" is a typo or placeholder, and the actual answer is in terms of t, as the problem mentions "how many 1 coins does she have?" So, unless there's more context, we can only express the answer in terms of t. But in the original problem statement, if t is given as "TNYWR," which we don't know what that stands for, maybe it's a number that was supposed to be written there but got replaced with "TNYWR" by mistake. Since the user is providing the problem, and the instruction is to answer it, perhaps they made a typo. Alternatively, maybe "TNYWR" is supposed to be an acronym for something else. Wait, TNYWR could stand for "The Next Year's World Record" or something, but that's not helpful here. Alternatively, maybe "TNYWR" is a code for a number? But without more context, it's hard to say. Wait, but in the problem statement, the user wrote: "Let t be TNYWR. Gray has t dollars..." So maybe "TNYWR" is a label or a specific value that was supposed to be defined but wasn't. However, given that the problem is presented in this way, perhaps we are to answer in terms of t. Because otherwise, if TNYWR is a number, we need to know what it is. But since it's not defined here, perhaps the problem is to express the number of 1 coins in terms of t, which is what we did earlier: x = t / 3. So, considering that, the answer should be t divided by 3, so the number of 1 coins is t/3. But since the number of coins has to be an integer, t must be a multiple of 3. However, unless the problem specifies constraints on t, we can just state the answer as t over 3. Let me check once again.Suppose t is 6 dollars. Then, according to this, x would be 2. So, 2 1 coins and 2 2 coins: 2*1 + 2*2 = 2 + 4 = 6. That works. If t is 12, then x would be 4: 4*1 + 4*2 = 4 + 8 = 12. That also works. If t is 3, then x is 1: 1*1 + 1*2 = 3. That's correct. So, the formula seems to hold. Therefore, unless there's a restriction I'm missing, the answer should be t divided by 3. Therefore, the number of 1 coins is t/3. So, the final answer is boxed{dfrac{t}{3}}.Wait, but the problem says "how many 1 coins does she have?" If t is not a multiple of 3, then the answer would not be an integer. But the problem must have a valid answer, so perhaps t is given as a multiple of 3. However, since t is defined as "TNYWR," which we don't know the value of, maybe the problem is expecting an answer in terms of t, as we did. Alternatively, maybe "TNYWR" is a typo, and the original problem had a specific number. For example, maybe it was supposed to be "Let t be 12" or some number, and "TNYWR" is a placeholder. Since we can't know, but in the problem as given, t is a variable, so the answer should be in terms of t. Therefore, the answer is t over 3. Therefore, I think the answer is boxed{dfrac{t}{3}}.</think>Gray has ( t ) dollars consisting of 1 and 2 coins, with the same number of each type of coin. Let ( x ) be the number of 1 coins. Since the number of 2 coins is also ( x ), the total amount of money can be expressed as:[1 cdot x + 2 cdot x = 3x]Given that this total amount is ( t ) dollars, we have:[3x = t]Solving for ( x ), the number of 1 coins, we get:[x = frac{t}{3}]Thus, the number of 1 coins Gray has is (boxed{dfrac{t}{3}}).,<think>Okay, so Gray has t dollars, and this amount is made up of 1 and 2 coins. The key point here is that she has the same number of both types of coins. Let me try to break this down step by step.First, let's define some variables. Let's say the number of 1 coins she has is x. Since she has the same number of 2 coins, that would also be x. So total coins of each type are equal. Now, the total amount of money from the 1 coins would be 1 times x, which is just x dollars. Similarly, the total from the 2 coins would be 2 times x, which is 2x dollars. If we add those two amounts together, that should equal the total amount of money Gray has, which is t dollars. So the equation would be:x (from 1 coins) + 2x (from 2 coins) = tCombining those terms, that's 3x = t.To find the number of 1 coins, which is x, we can solve for x by dividing both sides by 3. So x = t / 3.Wait, but hold on. The problem says that t is TNYWR. Hmm, I'm not sure what TNYWR stands for here. Maybe it's a typo or a placeholder? The original problem mentions "Let t be TNYWR." Maybe that's irrelevant, like maybe TNYWR is an abbreviation for something else, but in the context of the problem, since they're asking for the number of 1 coins in terms of t, maybe we can just proceed with the equation we have.So according to the equation, x is equal to t divided by 3. Therefore, the number of 1 coins she has is t over 3. But let me double-check. If there are x 1 coins and x 2 coins, the total is x + 2x = 3x. So 3x = t. Solving for x gives x = t/3. Yeah, that makes sense. For example, if t was 3, then she would have 1 of each coin, totaling 1*1 + 1*2 = 3, which works. If t was 6, then x would be 2, so 2 1 coins and 2 2 coins, totaling 2 + 4 = 6. That checks out.Therefore, the answer should be t divided by 3. Since the question asks for the number of 1 coins, which is x, that's x = t/3. So the answer is t over three. But since they might want it in boxed notation, probably boxed as boxed{dfrac{t}{3}}.</think>Gray has ( t ) dollars consisting of 1 and 2 coins, with the same number of each type. Let ( x ) be the number of 1 coins (and also the number of 2 coins). The total amount of money from the 1 coins is ( x ) dollars, and from the 2 coins is ( 2x ) dollars. Therefore, the equation is:[x + 2x = t]Simplifying this, we get:[3x = t]Solving for ( x ):[x = frac{t}{3}]Thus, the number of 1 coins Gray has is (boxed{dfrac{t}{3}}).]

🔑:Let k be the number of 1 coins that Gray has. It is given that k is also the number of 2 coins, which means the total amount of money in dollars that Gray has isk(1)+k(2)=k(1+2)=k(3) .The answer to (a) is 12 . Therefore, 3 k= t so k=frac{t}{3}=frac{12}{3}=4.