🧠:[<think>Okay, let's see. So we have this problem here. Let me read it again and try to parse it step by step.The problem says: a is a real number. x_1 and x_2 are the distinct roots of the equation x^2 + ax + 2 = x. Then, x_3 and x_4 are the distinct roots of the equation (x - a)^2 + a(x - a) + 2 = x. It's given that x_3 - x_1 = 3(x_4 - x_2). We need to find what x_4 - x_2 will be, with the options given.Alright, so first, maybe I need to rewrite both quadratic equations in standard form and find expressions for their roots. Then use Vieta's formulas or something related to the relationships between the roots. Then set up the equation given (x3 - x1 = 3(x4 - x2)) and solve for x4 - x2. Let's start with the first equation.First equation: x^2 + a x + 2 = xLet me rearrange that to standard quadratic form:x^2 + a x + 2 - x = 0Combine like terms:x^2 + (a - 1)x + 2 = 0So the quadratic is x^2 + (a - 1)x + 2 = 0. Let's denote this as equation (1).The roots of equation (1) are x1 and x2. Since they are distinct, the discriminant must be positive. Let me check that real quick. The discriminant D1 is [ (a - 1)^2 - 8 ]. For distinct roots, D1 > 0, so (a - 1)^2 > 8. But maybe we don't need to worry about that right now.Similarly, the second equation: (x - a)^2 + a(x - a) + 2 = xLet me expand this step by step.First, expand (x - a)^2 which is x^2 - 2a x + a^2.Then add a(x - a) which is a x - a^2.So combining these terms:x^2 - 2a x + a^2 + a x - a^2 + 2 = xSimplify the terms:- The a^2 and -a^2 cancel out.- The x terms: -2a x + a x = -a xSo now we have:x^2 - a x + 2 = xThen subtract x from both sides to bring everything to the left:x^2 - a x + 2 - x = 0Combine like terms:x^2 - (a + 1)x + 2 = 0So the second quadratic equation is x^2 - (a + 1)x + 2 = 0. Let's denote this as equation (2). The roots are x3 and x4. Again, since they are distinct, discriminant D2 must be positive: [ (a + 1)^2 - 8 ] > 0. So (a + 1)^2 > 8. Again, maybe not immediately important.So now, we have equation (1): x² + (a - 1)x + 2 = 0 with roots x1, x2.Equation (2): x² - (a + 1)x + 2 = 0 with roots x3, x4.We need to relate x3 - x1 = 3(x4 - x2). Find x4 - x2.Hmm. Let's think about Vieta's formulas here. For equation (1):Sum of roots: x1 + x2 = -(a - 1) = 1 - aProduct of roots: x1 x2 = 2For equation (2):Sum of roots: x3 + x4 = (a + 1)Product of roots: x3 x4 = 2So we have these relationships. Now, the problem gives us that x3 - x1 = 3(x4 - x2). Let's write this as:x3 - x1 = 3x4 - 3x2Rearranging:x3 - 3x4 = x1 - 3x2Hmm, not sure if that helps. Alternatively, maybe express x3 in terms of x1 and x4 in terms of x2.But perhaps we need to find expressions for x3 and x4 in terms of the coefficients, or maybe relate the roots of equation (2) to those of equation (1). Alternatively, maybe express x3 and x4 in terms of x1 and x2? Let's see.Alternatively, perhaps we can find expressions for x3 - x1 and x4 - x2, then set up the equation.But how?Alternatively, since the products of both equations are the same (2). Interesting. So x1 x2 = 2 and x3 x4 = 2. Maybe that's useful.Alternatively, perhaps express x3 and x4 in terms of equation (2), and x1 and x2 in terms of equation (1). Then, set up the equation x3 - x1 = 3(x4 - x2). Let's see.Alternatively, maybe use substitution. Let me think. The second equation is in terms of (x - a). Wait, the original problem says the second equation is (x - a)^2 + a(x - a) + 2 = x. Wait, maybe there's a substitution here. If I let y = x - a, then the equation becomes y² + a y + 2 = x. But x = y + a, so substitute back:y² + a y + 2 = y + aThen, bring all terms to left-hand side:y² + a y + 2 - y - a = 0Combine like terms:y² + (a - 1)y + (2 - a) = 0So in terms of y, the equation is y² + (a - 1)y + (2 - a) = 0. Hmm, interesting. But equation (1) is x² + (a - 1)x + 2 = 0. So the equations are similar but not the same. The constant term here is 2 - a, while in equation (1) it's 2.But maybe this substitution is helpful? The roots of this equation in y are x3 - a and x4 - a, since y = x - a, so x = y + a. Therefore, the roots y3 and y4 of the equation y² + (a - 1)y + (2 - a) = 0 correspond to y3 = x3 - a and y4 = x4 - a. So maybe we can relate the roots of equation (1) and equation (2) via this substitution.Alternatively, perhaps not. Let's see.Alternatively, perhaps let's write both quadratics and see their relations.Equation (1): x² + (a - 1)x + 2 = 0Equation (2): x² - (a + 1)x + 2 = 0So the two quadratics have the same constant term (2) but different coefficients for x. Let's denote equation (1) as:f(x) = x² + (a - 1)x + 2Equation (2) as:g(x) = x² - (a + 1)x + 2So, we can note that the sum of the roots for equation (1) is (1 - a) and for equation (2) is (a + 1). The product of roots is the same, 2.But how does this help us?We need to relate x3 - x1 = 3(x4 - x2). Let me think. Maybe express x3 and x4 in terms of the sum and product, and x1 and x2 similarly, then find the difference.Alternatively, let's note that x3 and x4 are roots of equation (2), so:x3 + x4 = a + 1x3 x4 = 2Similarly, x1 + x2 = 1 - ax1 x2 = 2So, perhaps, if we can write x3 in terms of x1 and x4 in terms of x2, or something like that. Alternatively, maybe express x3 - x1 and x4 - x2 in terms of a.But how?Alternatively, let's suppose that x3 - x1 = 3(x4 - x2). Let's call this equation (3). Then, we can write x3 = 3x4 - 3x2 + x1. Then, we can use the relationships from Vieta's formulas.But let's see:We know that x3 + x4 = a + 1, so x3 = (a + 1) - x4. Similarly, x1 + x2 = 1 - a, so x1 = (1 - a) - x2.Substituting these into equation (3):[(a + 1) - x4] - [(1 - a) - x2] = 3(x4 - x2)Simplify left side:(a + 1) - x4 - 1 + a + x2 = 3x4 - 3x2Combine like terms:(a + 1 - 1 + a) + (-x4 + x2) = 3x4 - 3x2So:2a + (-x4 + x2) = 3x4 - 3x2Bring all terms to left side:2a - x4 + x2 - 3x4 + 3x2 = 0Combine like terms:2a -4x4 +4x2 = 0Divide through by 4:(2a)/4 - x4 + x2 = 0 => (a/2) - x4 + x2 = 0Rearrange:x2 - x4 = -a/2Multiply both sides by -1:x4 - x2 = a/2So, x4 - x2 = a/2. So, the answer is (A) a/2. Wait, but let me check my steps again to make sure I didn't make a mistake.Starting from:x3 - x1 = 3(x4 - x2)Expressed x3 = (a + 1) - x4 and x1 = (1 - a) - x2. Then substituted in:[(a + 1) - x4] - [(1 - a) - x2] = 3(x4 - x2)Left side:(a + 1 - 1 + a) + (-x4 + x2) = 2a + (-x4 + x2)Set equal to 3x4 - 3x2So:2a - x4 + x2 = 3x4 - 3x2Bring all terms to left:2a - x4 + x2 -3x4 +3x2 = 0Combine:2a -4x4 +4x2 =0Factor:2a +4(x2 -x4)=0Divide by 2:a +2(x2 -x4)=0Then:2(x2 -x4)= -aSo:x2 -x4 = -a/2Multiply by -1:x4 -x2= a/2Yes, that seems correct. Therefore, x4 -x2 is a/2, which is option A. But the options are given as (A) a/2, (B) a/3, etc. So the answer should be A. Hmm, but let me check again if all steps are valid.Wait, so starting from x3 - x1 = 3(x4 - x2). Then, x3 = 3(x4 - x2) + x1. Then, using Vieta's formula for the sum x3 + x4 = a + 1, so substitute x3:3(x4 - x2) + x1 + x4 = a + 1Then, rearrange:4x4 -3x2 +x1 = a +1But we also have from Vieta's formula for equation (1): x1 + x2 =1 -a. So x1 =1 -a -x2. Substitute x1 into the above equation:4x4 -3x2 +1 -a -x2 =a +1Simplify:4x4 -4x2 +1 -a =a +1Bring 1 to both sides:4x4 -4x2 -a =aThen:4x4 -4x2 =2aDivide both sides by 4:x4 -x2 = (2a)/4 = a/2Same result. So that's correct. Therefore, the answer is a/2, which is option A.But let me check the problem statement again. The options are:(A) a/2(B) a/3(C) 2a/3(D) 3a/2(E) NoneSo according to our solution, the answer is A. But maybe the test-makers want to trick us? Let me verify with a specific example.Suppose we pick a value for a and check. Let's choose a = 2.First, equation (1): x² + (2 -1)x +2 = x² +x +2 =0. Wait, original equation was x² + a x +2 =x. So a=2, equation is x² +2x +2 =x => x² +x +2=0. The roots x1 and x2. Let's compute discriminant: 1 -8 = -7. Wait, but the roots would be complex. But the problem states that x1 and x2 are distinct real roots. So discriminant must be positive. So a must satisfy (a -1)^2 -8 >0. For a=2, (2-1)^2 -8=1-8=-7 <0. So complex roots. So a=2 is invalid.So let's pick a value of a such that both discriminants are positive. For equation (1): (a -1)^2 >8 => a-1 > sqrt(8) or a -1 < -sqrt(8) => a >1 + 2√2 or a <1 -2√2. Similarly, equation (2): (a +1)^2 >8 => a +1 > 2√2 or a +1 < -2√2 => a > -1 +2√2 ≈1.828 or a < -1 -2√2≈-3.828. So for example, let's pick a=4. Then check:Equation (1): x² + (4 -1)x +2 =x² +3x +2=0. Wait, that factors as (x+1)(x+2)=0. So roots x1=-1, x2=-2.Equation (2): (x -4)^2 +4(x -4) +2 =x. Let's compute:(x -4)^2 =x² -8x +164(x -4)=4x -16So adding them up: x² -8x +16 +4x -16 +2 =x² -4x +2 =xBring x to left: x² -5x +2=0So equation (2) is x² -5x +2=0. Compute roots x3 and x4 using quadratic formula:x=(5 ±sqrt(25 -8))/2=(5±sqrt(17))/2. So approximate roots:x3=(5 +sqrt(17))/2≈(5+4.123)/2≈4.5615x4=(5 -sqrt(17))/2≈(5-4.123)/2≈0.4385Now compute x3 -x1≈4.5615 - (-1)=5.5615x4 -x2≈0.4385 - (-2)=2.4385Check if x3 -x1 ≈3*(x4 -x2): 3*2.4385≈7.3155, which is not equal to 5.5615. So the equation doesn't hold. But according to our solution, x4 -x2 should be a/2=4/2=2. But here, x4 -x2≈2.4385, which is not 2. So something is wrong. Which contradicts our previous conclusion. Therefore, our solution must be incorrect.Wait, so there's a mistake. So our algebraic manipulation gives x4 -x2=a/2, but when we test with a=4, it's not matching. Therefore, our mistake must be in the algebraic steps. Let's check again.Let me go through the steps again carefully.Starting with x3 - x1 = 3(x4 -x2)We have Vieta's:x1 +x2 =1 -a (from equation 1)x3 +x4 =a +1 (from equation 2)So, from equation (1): x1 =1 -a -x2From equation (2): x3 =a +1 -x4Substitute into the given equation:x3 -x1 =3(x4 -x2)=> (a +1 -x4) - (1 -a -x2) = 3x4 -3x2Simplify left side:a +1 -x4 -1 +a +x2 =2a -x4 +x2So:2a -x4 +x2 =3x4 -3x2Bring all terms to left:2a -x4 +x2 -3x4 +3x2 =0Combine like terms:2a -4x4 +4x2=0Divide by 2:a -2x4 +2x2=0Rearranged:a +2x2 -2x4=0But wait, here's a mistake. Wait, original equation after combining:2a -4x4 +4x2=0If we factor 2:2(a -2x4 +2x2)=0 => a -2x4 +2x2=0So, a +2x2 -2x4=0Then, moving terms:2x2 -2x4 = -aDivide both sides by 2:x2 -x4 = -a/2Multiply by -1:x4 -x2 = a/2So, same as before. So according to this, x4 -x2 =a/2. However, when we tested with a=4, which should make x4 -x2=2, but in reality, when a=4, x4 -x2≈0.4385 -(-2)=2.4385≠2. So inconsistency.Thus, our algebraic solution must be incorrect. There's a mistake in assuming the given condition holds for all a, but perhaps the given condition x3 -x1=3(x4 -x2) imposes a specific value on a, which we need to solve for, and then compute x4 -x2. Wait, the problem says "If x3 -x1=3(x4 -x2), then x4 -x2 will be?" So maybe x4 -x2 is not simply a/2, but we need to find it in terms of a, but perhaps a is determined by the condition. Wait, but the answer choices are in terms of a, so the answer is supposed to be in terms of a, not a specific number. So maybe the answer is a/2. But in our test case, with a=4, the answer was approximately 2.4385, which is a/2=2, but 2.4385≈2.44≠2. So there's a contradiction.Thus, our initial approach must have an error. Let's re-examine.Wait, perhaps the mistake is that when we substituted Vieta's formula, we treated the roots as variables, but maybe we need to consider the specific relations between the quadratics. Let's try another approach.Given that both quadratics have the same product of roots, which is 2. So x1 x2 =x3 x4=2.Additionally, equation (1): x1 +x2=1 -aEquation (2):x3 +x4=a +1Given x3 -x1=3(x4 -x2). Let's denote this as equation (3).We need to solve for x4 -x2.Let me consider variables x1, x2, x3, x4 with the above conditions.Let me define S1 = x1 +x2=1 -aS2 =x3 +x4=a +1P= x1x2=x3x4=2Let me denote D1= x1 -x2, and D2= x3 -x4. Wait, but maybe not necessary.Alternatively, let's express x3 in terms of x4: x3=S2 -x4= a +1 -x4Similarly, x1= S1 -x2=1 -a -x2Plug into equation (3):x3 -x1=3(x4 -x2)Substitute x3 and x1:(a +1 -x4) - (1 -a -x2)=3x4 -3x2Simplify left side:a +1 -x4 -1 +a +x2=2a -x4 +x2So equation becomes:2a -x4 +x2=3x4 -3x2Bring all terms to left:2a -x4 +x2 -3x4 +3x2=0Combine like terms:2a -4x4 +4x2=0Then:4x2 -4x4= -2aDivide by 4:x2 -x4= -a/2Multiply by -1:x4 -x2= a/2Same result as before. But in the test case with a=4, the actual value was 2.4385≈a/2=2, which is conflicting.But in our test case, when a=4, equation (1) had roots x1=-1, x2=-2 (sum=-3=1 -a=1 -4=-3, product=2). Equation (2) had roots x3≈4.5615, x4≈0.4385 (sum=5=a +1=4 +1=5, product≈2). Then x3 -x1≈4.5615 -(-1)=5.5615, and x4 -x2≈0.4385 -(-2)=2.4385. Then 3*(x4 -x2)=7.3155. But 5.5615≈7.3155? No, not equal. Therefore, for a=4, the given condition x3 -x1=3(x4 -x2) does not hold, which is why when we forced a=4, it's not satisfying the condition. Therefore, the value of a is not arbitrary. The problem states that a is a real number, and given the condition x3 -x1=3(x4 -x2), we have to find x4 -x2. So actually, a is fixed by the condition. Wait, but the answer is given in terms of a, so a must be a variable, and the answer is expressed in terms of a. But in the problem, it's possible that such a condition can only hold for certain a, and in that case, x4 -x2 is a specific multiple of a. Wait, but how?Wait, maybe we need to solve for a first. Let's try.We have from the previous steps:x4 -x2 =a/2. So x4 -x2 is a/2. However, in our test case, when a=4, x4 -x2 was ≈2.44, which is a/2=2. So discrepancy here. Therefore, in reality, when the condition x3 -x1=3(x4 -x2) is imposed, it actually constrains the value of a. So we need to find a such that x4 -x2=a/2 and x3 -x1=3*(x4 -x2). But how?Wait, but if we already derived x4 -x2= a/2 under the given condition, then regardless of the test case, which didn't satisfy the condition, the answer would still be a/2. Because in the problem, it's given that x3 -x1=3(x4 -x2). So under that condition, x4 -x2 must equal a/2, hence answer is A. However, in our test case, we picked a=4, which didn't satisfy the condition, so the discrepancy arises because that a=4 is not compatible with the condition. So perhaps when we choose a such that the condition holds, x4 -x2 will indeed be a/2.Therefore, maybe the answer is indeed A. Let me verify with another approach.Alternatively, we can write the equations as:From equation (1): x1 +x2=1 -a and x1x2=2From equation (2): x3 +x4=a +1 and x3x4=2Given that x3 -x1=3(x4 -x2). Let's denote d =x4 -x2, then x3 -x1=3d. So x3 =x1 +3d and x4=x2 +d.Now, since x3 +x4= a +1, substitute:x1 +3d +x2 +d= a +1 => (x1 +x2) +4d= a +1But x1 +x2=1 -a, so:1 -a +4d =a +1Simplify:1 -a +4d =a +1Subtract 1 from both sides: -a +4d =aBring -a to the right:4d=2a => d= (2a)/4= a/2Therefore, d= x4 -x2= a/2. So that's the same result. Therefore, the answer is A.Therefore, the test case I did earlier was invalid because I chose a=4 arbitrarily, which doesn't satisfy the given condition. So in reality, when the condition is imposed, a is such that x4 -x2= a/2. Therefore, answer is A. So despite the test case not working, the algebraic manipulation under the given condition shows that x4 -x2= a/2. Therefore, the answer should be A.But wait, the answer choices include "None". So why the test case didn't satisfy? Because in the test case, the condition isn't met. Therefore, for values of a where the condition is met, x4 -x2= a/2. Therefore, if such a exists, answer is A, otherwise, it's E. But the problem states "a is a real number. x1 and x2 are the distinct roots... x3 and x4 are the distinct roots... If x3 -x1=3(x4 -x2), then x4 -x2 will be?" So as long as such a exists, which the problem implies by giving the options including A to D, then the answer is A. So the answer is A. Therefore, I think the answer is A.Final Answerboxed{A}</think>Given the problem, we need to find the value of ( x_4 - x_2 ) under the condition ( x_3 - x_1 = 3(x_4 - x_2) ). First, we rewrite the given quadratic equations in standard form:1. The equation ( x^2 + ax + 2 = x ) becomes: [ x^2 + (a - 1)x + 2 = 0 ] The roots ( x_1 ) and ( x_2 ) satisfy: [ x_1 + x_2 = 1 - a quad text{and} quad x_1 x_2 = 2 ]2. The equation ( (x - a)^2 + a(x - a) + 2 = x ) becomes: [ x^2 - (a + 1)x + 2 = 0 ] The roots ( x_3 ) and ( x_4 ) satisfy: [ x_3 + x_4 = a + 1 quad text{and} quad x_3 x_4 = 2 ]Given the condition ( x_3 - x_1 = 3(x_4 - x_2) ), we start by expressing ( x_3 ) and ( x_1 ) in terms of ( x_4 ) and ( x_2 ):- From the sum of the roots, we have ( x_3 = (a + 1) - x_4 ) and ( x_1 = (1 - a) - x_2 ).Substituting these into the given condition:[(a + 1 - x_4) - (1 - a - x_2) = 3(x_4 - x_2)]Simplifying the left-hand side:[a + 1 - x_4 - 1 + a + x_2 = 2a - x_4 + x_2]Equating this to the right-hand side:[2a - x_4 + x_2 = 3x_4 - 3x_2]Rearranging terms:[2a - x_4 + x_2 - 3x_4 + 3x_2 = 0]Combining like terms:[2a - 4x_4 + 4x_2 = 0]Solving for ( x_4 - x_2 ):[4x_2 - 4x_4 = -2a implies x_2 - x_4 = -frac{a}{2} implies x_4 - x_2 = frac{a}{2}]Thus, the value of ( x_4 - x_2 ) is ( boxed{A} ).,<think>Okay, so I have this problem here. Let me try to parse it step by step. The problem says:a is a real number. x_1 and x_2 are the distinct roots of x^2 + ax + 2 = x. x_3 and x_4 are the distinct roots of (x - a)^2 + a(x - a) + 2 = x. If x_3 - x_1 = 3(x_4 - x_2), then x_4 - x_2 will be? The options are given as (A) to (E), with (E) being "None".Alright, first, let's try to rewrite the equations given for both quadratics. For the first equation: x^2 + ax + 2 = x. Let's rearrange that to standard quadratic form. Subtract x from both sides:x^2 + ax + 2 - x = 0Combine like terms:x^2 + (a - 1)x + 2 = 0So, the first quadratic equation is x^2 + (a - 1)x + 2 = 0, with roots x_1 and x_2. Since they are distinct roots, the discriminant must be positive. But maybe we don't need that right now.Similarly, the second equation is (x - a)^2 + a(x - a) + 2 = x. Let's expand this step by step.First, expand (x - a)^2:(x^2 - 2a x + a^2) + a(x - a) + 2 = xThen expand the a(x - a): x^2 - 2a x + a^2 + a x - a^2 + 2 = xCombine like terms:- For the x^2 term: just x^2- For the x terms: -2a x + a x = (-2a + a)x = -a x- For the constants: a^2 - a^2 = 0, and then +2So the left side simplifies to:x^2 - a x + 2 = xAgain, let's bring the x to the left side:x^2 - a x + 2 - x = 0Combine like terms:x^2 - (a + 1)x + 2 = 0Therefore, the second quadratic equation is x^2 - (a + 1)x + 2 = 0, with roots x_3 and x_4.Now, the problem states that x_3 - x_1 = 3(x_4 - x_2), and we need to find x_4 - x_2.Let me note down the two quadratic equations:1. First equation: x^2 + (a - 1)x + 2 = 0 (roots x_1, x_2)2. Second equation: x^2 - (a + 1)x + 2 = 0 (roots x_3, x_4)Let me recall Vieta's formulas for quadratic equations. For a quadratic equation x^2 + bx + c = 0, the sum of roots is -b and the product is c.So for the first equation:Sum of roots: x_1 + x_2 = -(a - 1) = 1 - aProduct of roots: x_1 x_2 = 2For the second equation:Sum of roots: x_3 + x_4 = (a + 1)Product of roots: x_3 x_4 = 2Interesting, the product of roots is the same for both quadratics, which is 2.Now, we need to relate x_3 - x_1 and x_4 - x_2. The given condition is x_3 - x_1 = 3(x_4 - x_2).Let me denote the differences. Let me think if we can express these differences in terms of the roots' sums and products.Alternatively, maybe express the roots in terms of the coefficients and then compute the differences.Alternatively, let's denote variables for the roots:From first equation:x_1 + x_2 = 1 - aFrom second equation:x_3 + x_4 = a + 1We need to relate x_3 - x_1 and x_4 - x_2.Given x_3 - x_1 = 3(x_4 - x_2), let me rearrange this equation:x_3 - x_1 = 3x_4 - 3x_2Bring all terms to one side:x_3 - 3x_4 - x_1 + 3x_2 = 0But not sure if that helps. Alternatively, express x_3 in terms of x_1 and x_4, x_2:x_3 = 3x_4 - 3x_2 + x_1But maybe not directly helpful.Alternatively, perhaps express the roots in terms of each other. Let me think.Alternatively, notice that both quadratics have the same constant term (product of roots is 2). Maybe there's a relationship between the roots.But since the products are the same, perhaps x_3 and x_4 are related to x_1 and x_2 in some reciprocal way? Not sure. Let me check.Wait, the product is the same, but the sum is different. The first quadratic has sum 1 - a, the second has sum a + 1.Alternatively, perhaps relate the roots of the second equation to the roots of the first equation shifted by some amount. Because the second equation is in terms of (x - a), which suggests a substitution.Wait, let me look back at the original second equation: (x - a)^2 + a(x - a) + 2 = xIs there a substitution that can relate this equation to the first equation?Let me let y = x - a, then the equation becomes:y^2 + a y + 2 = xBut since y = x - a, then x = y + a, so substitute back:y^2 + a y + 2 = y + aBring all terms to the left:y^2 + a y + 2 - y - a = 0Combine like terms:y^2 + (a - 1)y + (2 - a) = 0Hmm, interesting. So this equation in terms of y is:y^2 + (a - 1)y + (2 - a) = 0Compare this to the first equation, which was:Original first equation: x^2 + (a - 1)x + 2 = 0So if we compare the two equations:First equation (in x): x^2 + (a - 1)x + 2 = 0Second equation (in y): y^2 + (a - 1)y + (2 - a) = 0So they have the same coefficients for the quadratic and linear terms, but different constant terms.Therefore, the roots of the second equation (in y) are different from the first equation (in x). Therefore, if we denote the roots of the second equation (in y) as y_1 and y_2, then y_1 + y_2 = -(a - 1) and y_1 y_2 = 2 - a.But since y = x - a, then the roots in terms of x are x_3 = y_1 + a and x_4 = y_2 + a.Therefore, we can relate the roots x_3 and x_4 to the roots of this transformed equation. Let me formalize this.Let me denote the roots of the first equation (in x) as x_1 and x_2, so:x_1 + x_2 = 1 - a and x_1 x_2 = 2.For the second equation, in terms of y, the roots y_1 and y_2 satisfy:y_1 + y_2 = -(a - 1) = 1 - ay_1 y_2 = 2 - aThen, since x_3 = y_1 + a and x_4 = y_2 + a, then:x_3 + x_4 = (y_1 + a) + (y_2 + a) = (y_1 + y_2) + 2a = (1 - a) + 2a = 1 + aWhich matches the earlier Vieta's sum.Similarly, x_3 x_4 = (y_1 + a)(y_2 + a) = y_1 y_2 + a(y_1 + y_2) + a^2 = (2 - a) + a(1 - a) + a^2 = 2 - a + a - a^2 + a^2 = 2Which matches the product.So, that's consistent.But how does this help us? Maybe we can express x_3 and x_4 in terms of y_1 and y_2, which are roots of an equation similar to the first one but with a different constant term.Alternatively, perhaps relate x_3 - x_1 and x_4 - x_2 through substitutions.Wait, the problem gives us x_3 - x_1 = 3(x_4 - x_2). Let me write this as:x_3 - x_1 = 3x_4 - 3x_2Let me rearrange terms:x_3 - 3x_4 = x_1 - 3x_2But not sure if this is helpful. Alternatively, maybe express x_3 and x_4 in terms of y_1 and y_2 and then relate to x_1 and x_2.Alternatively, since x_1 and x_2 are roots of the first equation, which is similar to the equation in y except for the constant term. The first equation (x) has constant term 2, and the equation in y has constant term 2 - a.Hmm. Maybe there's a relationship between y_1 and x_1, but not necessarily. Alternatively, if we think of the equations, the equation in y is y^2 + (a - 1)y + (2 - a) = 0, while the first equation is x^2 + (a - 1)x + 2 = 0. So, if we let x = y + k, maybe we can find a substitution. Let's see.Suppose x = y + k. Then, substitute into the first equation:(y + k)^2 + (a - 1)(y + k) + 2 = 0Expand:y^2 + 2k y + k^2 + (a - 1)y + (a - 1)k + 2 = 0Combine like terms:y^2 + [2k + (a - 1)] y + [k^2 + (a - 1)k + 2] = 0Compare this to the equation in y:y^2 + (a - 1)y + (2 - a) = 0If we want the coefficients to match:For the y term: 2k + (a - 1) = (a - 1) => 2k = 0 => k = 0But then the constant term would be k^2 + (a - 1)k + 2 = 0 + 0 + 2 = 2 ≠ 2 - a. So, no, unless a is 0, which it isn't necessarily. So this substitution doesn't help.Alternatively, maybe another substitution. But perhaps this is overcomplicating.Alternatively, let's use the given condition:x_3 - x_1 = 3(x_4 - x_2)Let me write this as:x_3 - x_1 = 3x_4 - 3x_2Then, rearranged:x_3 - 3x_4 = x_1 - 3x_2 --- (1)But I don't know if this helps. Alternatively, maybe express x_3 in terms of x_4 and substitute into the sum for the second equation.From Vieta's formula for the second equation:x_3 + x_4 = a + 1 --- (2)So, if we can express equation (1) in terms of (2), perhaps?From equation (1):x_3 = 3x_4 - 3x_2 + x_1Substitute into equation (2):3x_4 - 3x_2 + x_1 + x_4 = a + 1Combine terms:4x_4 - 3x_2 + x_1 = a + 1 --- (3)Now, from Vieta's formula for the first equation:x_1 + x_2 = 1 - a --- (4)So, from equation (4), we can express x_1 = 1 - a - x_2Substitute x_1 into equation (3):4x_4 - 3x_2 + (1 - a - x_2) = a + 1Simplify:4x_4 - 3x_2 + 1 - a - x_2 = a + 1Combine like terms:4x_4 - 4x_2 + 1 - a = a + 1Subtract 1 from both sides:4x_4 - 4x_2 - a = aBring the a to the right:4x_4 - 4x_2 = 2aFactor out 4:4(x_4 - x_2) = 2aDivide both sides by 2:2(x_4 - x_2) = aTherefore:x_4 - x_2 = a / 2But let me check the options. Option A is a/2, so that's a possibility. But let me verify my steps again to ensure I didn't make a mistake.So, starting from the given condition:x_3 - x_1 = 3(x_4 - x_2)Then, x_3 = 3x_4 - 3x_2 + x_1Substituted into Vieta's sum x_3 + x_4 = a + 1:3x_4 - 3x_2 + x_1 + x_4 = a + 1 => 4x_4 - 3x_2 + x_1 = a + 1Then, from Vieta's first equation x_1 + x_2 = 1 - a => x_1 = 1 - a - x_2Substitute x_1 into above equation:4x_4 - 3x_2 + 1 - a - x_2 = a + 1Combine x_2 terms: -3x_2 - x_2 = -4x_2Thus, 4x_4 -4x_2 +1 -a = a +1Subtract 1: 4x_4 -4x_2 -a = aThen, 4x_4 -4x_2 = 2aFactor: 4(x_4 -x_2) = 2a => 2(x_4 -x_2) = a => x_4 -x_2 = a/2Which is option A.But let me check if this answer is among the choices. Yes, option A is a/2. But let me verify once again because sometimes there might be an error in substitution.Alternatively, perhaps I can approach the problem by using the roots expressions.Let me denote the roots for the first quadratic equation:x^2 + (a -1)x + 2 =0The roots are x1 and x2. Using quadratic formula:x = [-(a -1) ± sqrt((a -1)^2 - 8)] / 2Similarly, for the second equation:x^2 - (a +1)x + 2 =0Roots x3 and x4:x = [(a +1) ± sqrt((a +1)^2 - 8)] / 2So, the roots are:First equation:x1 = [1 -a + sqrt((a -1)^2 -8)] / 2x2 = [1 -a - sqrt((a -1)^2 -8)] / 2Second equation:x3 = [a +1 + sqrt((a +1)^2 -8)] / 2x4 = [a +1 - sqrt((a +1)^2 -8)] / 2Now, the condition given is x3 - x1 = 3(x4 - x2). Let's compute x3 - x1 and x4 - x2.Compute x3 - x1:= [ (a +1) + sqrt((a +1)^2 -8) ) / 2 ] - [ (1 -a + sqrt((a -1)^2 -8)) / 2 ]= [ (a +1 -1 +a) + sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8) ) ] / 2Simplify numerator:= [ 2a + sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8) ] / 2Similarly, compute x4 - x2:= [ (a +1 - sqrt((a +1)^2 -8)) / 2 ] - [ (1 -a - sqrt((a -1)^2 -8)) / 2 ]= [ (a +1 -1 +a) - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8) ) ] / 2Simplify numerator:= [ 2a - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8) ] / 2Given that x3 - x1 = 3(x4 - x2):So,[ 2a + sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8) ] / 2 = 3 * [ 2a - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8) ] / 2Multiply both sides by 2 to eliminate denominators:2a + sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8) = 3[2a - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8)]Expand the right side:6a - 3 sqrt((a +1)^2 -8) + 3 sqrt((a -1)^2 -8)Now, bring all terms to the left side:2a + sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8) -6a + 3 sqrt((a +1)^2 -8) - 3 sqrt((a -1)^2 -8) = 0Combine like terms:(2a -6a) + [sqrt((a +1)^2 -8) +3 sqrt((a +1)^2 -8)] + [ - sqrt((a -1)^2 -8) -3 sqrt((a -1)^2 -8) ] = 0Simplify:-4a + 4 sqrt((a +1)^2 -8) -4 sqrt((a -1)^2 -8) = 0Divide both sides by -4:a - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8) = 0Rearranged:a = sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8)Let me denote:Let S = sqrt((a +1)^2 -8)Let T = sqrt((a -1)^2 -8)Then, the equation becomes:a = S - TSo, S = a + TNow, square both sides:S^2 = (a + T)^2But S^2 = (a +1)^2 -8Similarly, T^2 = (a -1)^2 -8So,(a +1)^2 -8 = a^2 + 2a T + T^2But T^2 = (a -1)^2 -8, so substitute:(a +1)^2 -8 = a^2 + 2a T + (a -1)^2 -8Simplify the equation:Left side: (a^2 + 2a +1) -8 = a^2 + 2a -7Right side: a^2 + 2a T + (a^2 -2a +1) -8 = 2a^2 -2a -7 + 2a TSet equal:a^2 + 2a -7 = 2a^2 -2a -7 + 2a TSubtract (a^2 + 2a -7) from both sides:0 = a^2 -4a + 2a TSo,a^2 -4a + 2a T =0Factor out a:a(a -4 + 2T)=0So, either a=0 or a -4 +2T=0Case 1: a=0Check if this is possible. If a=0, then check the original quadratics.First quadratic: x^2 +0x +2 =x => x^2 -x +2=0. The discriminant is (-1)^2 -8=1-8=-7, which is negative. But the problem states that there are distinct real roots. Therefore, a=0 is invalid because discriminant is negative, so no real roots. Thus, this case is rejected.Case 2: a -4 +2T=0 => 2T=4 -a => T=(4 -a)/2But T is defined as sqrt((a -1)^2 -8). So,sqrt((a -1)^2 -8) = (4 -a)/2Square both sides:(a -1)^2 -8 = (4 -a)^2 /4Multiply both sides by 4 to eliminate denominator:4(a -1)^2 -32 = (4 -a)^2Expand left side:4(a^2 -2a +1) -32 = 4a^2 -8a +4 -32 =4a^2 -8a -28Right side: (4 -a)^2 =16 -8a +a^2Set equal:4a^2 -8a -28 =16 -8a +a^2Subtract right side from left side:4a^2 -8a -28 -16 +8a -a^2=0 => 3a^2 -44=0Thus,3a^2 =44 => a^2=44/3 => a=±sqrt(44/3)=±(2*sqrt(33))/3But let's check if these values of a satisfy the original equation sqrt((a -1)^2 -8) = (4 -a)/2.Let me take a positive first: a=2*sqrt(33)/3 ≈ 2*5.7446/3≈3.8297Compute (4 -a)/2≈(4 -3.8297)/2≈0.1703/2≈0.08515Compute sqrt((a -1)^2 -8):a -1≈3.8297 -1≈2.8297(2.8297)^2≈8.00728.0072 -8≈0.0072, sqrt(0.0072)≈0.085, which matches (4 -a)/2≈0.08515. Close enough considering rounding. Similarly for the negative a=-2*sqrt(33)/3≈-3.8297Compute (4 -a)/2=(4 -(-3.8297))/2≈7.8297/2≈3.91485Compute sqrt((a -1)^2 -8):a -1≈-3.8297 -1≈-4.8297(-4.8297)^2≈23.32723.327 -8≈15.327, sqrt(15.327)≈3.915, which matches (4 -a)/2≈3.91485. So both roots are valid.Therefore, the possible values of a are ±2*sqrt(33)/3.Now, we need to compute x4 -x2. From the previous approach, we arrived at x4 -x2 =a/2. So, if a=±2*sqrt(33)/3, then x4 -x2=±sqrt(33)/3. However, the options given are in terms of a/2, a/3, etc. But if we have x4 -x2= a/2, then this would correspond to option A. However, the problem doesn't specify "in terms of a", but the answer choices are in terms of a. Wait, let me check the problem again.The question says: "If x_3 - x_1 = 3(x_4 - x_2), then x_4 - x_2 will be?"Options are:(A) a/2(B) a/3(C) 2a/3(D) 3a/2(E) NoneSo, according to my previous derivation, x4 -x2 =a/2, which is option A. However, when I checked using the quadratic formula, substituting specific values of a, we found that a is ±2*sqrt(33)/3, which would make x4 -x2= ±sqrt(33)/3. But the answer is supposed to be in terms of a. However, in the problem statement, a is a real number, and the answer choices are in terms of a, which suggests that x4 -x2 can be expressed in terms of a without needing to know the specific value of a. Therefore, my initial algebraic approach leading to x4 -x2 =a/2 would be correct, implying answer A.But why did the quadratic formula approach lead to specific a values? Because the condition given in the problem (x3 -x1=3(x4 -x2)) must hold for some specific a, but the question is asking for x4 -x2 in terms of a, given that the condition holds. Wait, but how can x4 -x2 be expressed purely in terms of a, if the condition itself relates a to the roots? Hmm.Wait, perhaps there's a misinterpretation here. Let's reconsider.The problem states that a is a real number. x1 and x2 are distinct roots of the first equation, x3 and x4 are distinct roots of the second equation. The condition given is x3 -x1 =3(x4 -x2). Then, we have to find x4 -x2.But in the problem, are we to find x4 -x2 in terms of a, given that the condition holds? But since the condition must hold for certain a, which we found to be ±2√33/3, then x4 -x2 would be a/2, but since a is specific, then x4 -x2 would be specific numbers. However, the answer choices are in terms of a. Therefore, possibly, the problem is expecting an expression in terms of a, regardless of the specific a that satisfies the condition, but in reality, the condition defines a relation that allows x4 -x2 to be expressed in terms of a, which is a/2, hence answer A.But according to the quadratic formula approach, when a is ±2√33/3, then x4 -x2=±√33/3, which is equal to a/2. Because a=2√33/3, so a/2=√33/3, and similarly for the negative. Therefore, even in that case, x4 -x2= a/2. So regardless of the value of a (as long as the condition holds), x4 -x2 is a/2. Hence, the answer is A.Therefore, despite the fact that a is fixed by the condition, x4 -x2 is a/2. Therefore, the answer should be A.But wait, in the initial approach, we found that x4 -x2= a/2, regardless of the value of a (as long as the equations have distinct roots and the condition holds). Therefore, the answer is A.Alternatively, maybe there's an error in assuming that the answer can be given in terms of a without considering the specific a's that satisfy the condition. However, the problem states that a is a real number, and given that the equations have distinct roots (so discriminants are positive), and that the condition x3 -x1=3(x4 -x2) holds. But the answer is requested as x4 -x2=? in terms of a, given that the condition holds. Therefore, since we derived algebraically that x4 -x2= a/2 under the given condition, the answer is A.But let me check the discriminant conditions.For the first equation: x² + (a -1)x +2=0. The discriminant is (a -1)^2 -8. For distinct real roots, this must be positive: (a -1)^2 >8 => |a -1| > 2√2.Similarly, the second equation: x² - (a +1)x +2=0. Discriminant is (a +1)^2 -8. For distinct roots, (a +1)^2 >8 => |a +1| >2√2.So, a must satisfy both |a -1|>2√2 and |a +1|>2√2.In our previous solutions, a=±2√33/3≈±3.8297. Let's check if these values satisfy the discriminant conditions.For a=2√33/3≈3.8297:|a -1|=|3.8297 -1|=2.8297≈2.83. 2√2≈2.828, so 2.83>2.828, so satisfies.|a +1|=|3.8297 +1|=4.8297>2.828, satisfies.Similarly for a=-2√33/3≈-3.8297:|a -1|=|-3.8297 -1|=4.8297>2.828|a +1|=|-3.8297 +1|=2.8297>2.828, so both satisfy.Therefore, these are valid values.But regardless of that, in the problem, the answer is given in terms of a, so regardless of the specific a (as long as the conditions hold), x4 -x2= a/2. Hence, the answer is A.Wait, but according to my first approach, using Vieta's formula and substitutions, I arrived directly at x4 -x2= a/2, which seems to be correct. The quadratic formula approach confirmed that when a is such that the condition holds, x4 -x2= a/2. Therefore, answer A is correct.But the answer options include "None". So, why is "None" an option? Perhaps there's a mistake in my reasoning.Wait, let me cross-verify once more.From the first approach, using Vieta's formula and substitution, I derived x4 -x2= a/2. Let's check if this holds with the specific a values we found.If a=2√33/3, then x4 -x2= a/2=√33/3. Let's compute x4 -x2 using the quadratic formula expressions.For the first equation, roots x1 and x2:x1 = [1 -a + sqrt((a -1)^2 -8)] / 2x2 = [1 -a - sqrt((a -1)^2 -8)] / 2Thus, x4 -x2 (wait, x4 is a root of the second equation. Let me get x4 and x2.x4 is a root of the second equation: x^2 - (a +1)x +2=0. So,x4 = [ (a +1) - sqrt((a +1)^2 -8) ] / 2x2 is [1 -a - sqrt((a -1)^2 -8)] / 2So,x4 -x2 = [ (a +1) - sqrt((a +1)^2 -8) ] / 2 - [1 -a - sqrt((a -1)^2 -8)] / 2Combine terms:= [ (a +1) - sqrt((a +1)^2 -8) -1 +a + sqrt((a -1)^2 -8) ] / 2Simplify numerator:= (2a) - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8)Thus,x4 -x2 = [2a - sqrt((a +1)^2 -8) + sqrt((a -1)^2 -8) ] / 2But from the condition in the problem, we had that:From the quadratic formula approach, we found that:a = sqrt((a +1)^2 -8) - sqrt((a -1)^2 -8)So, sqrt((a +1)^2 -8) = a + sqrt((a -1)^2 -8)Substitute into the expression for x4 -x2:x4 -x2 = [2a - (a + sqrt((a -1)^2 -8)) + sqrt((a -1)^2 -8) ] / 2Simplify numerator:2a -a - sqrt(...) + sqrt(...) = aThus,x4 -x2 = a / 2Which matches the previous result. Therefore, even with the quadratic formula approach, we confirm that x4 -x2 = a/2. Therefore, the answer is A.But the problem's answer choices include (A) a/2, which is exactly our conclusion. Therefore, unless there is a miscalculation somewhere, the answer should be A.But why is option E there? Maybe if the answer is not among the options, but according to this, A is correct.Alternatively, perhaps I made a mistake in interpreting the problem.Wait, let me check the original problem again:The first equation is x² + a x +2 =x. Which we rearranged to x² + (a -1)x +2 =0. Correct.The second equation is (x -a)^2 +a(x -a) +2 =x. Let's re-express this step by step to make sure:(x -a)^2 =x² -2a x +a²a(x -a) =a x -a²Thus, adding them and +2:x² -2a x +a² +a x -a² +2 = x² -a x +2So, equation becomes x² -a x +2 =x. Then moving x to left:x² - (a +1)x +2 =0. Correct.Thus, the second equation is x² - (a +1)x +2 =0. Correct.Then, Vieta's formulas are correct. Then, using the condition x3 -x1 =3(x4 -x2), through substitution we arrive at x4 -x2 =a/2. Quadratic formula approach confirmed this. Therefore, answer A is correct.Therefore, the answer is boxed{A}.Final Answerboxed{A}</think>Given the problem, we need to find the value of ( x_4 - x_2 ) given the condition ( x_3 - x_1 = 3(x_4 - x_2) ).First, we rewrite the given quadratic equations in standard form:1. The equation ( x^2 + ax + 2 = x ) becomes ( x^2 + (a - 1)x + 2 = 0 ). The roots ( x_1 ) and ( x_2 ) satisfy: - Sum: ( x_1 + x_2 = 1 - a ) - Product: ( x_1 x_2 = 2 )2. The equation ( (x - a)^2 + a(x - a) + 2 = x ) becomes ( x^2 - (a + 1)x + 2 = 0 ). The roots ( x_3 ) and ( x_4 ) satisfy: - Sum: ( x_3 + x_4 = a + 1 ) - Product: ( x_3 x_4 = 2 )Using Vieta's formulas and the given condition ( x_3 - x_1 = 3(x_4 - x_2) ), we start by expressing ( x_3 ) in terms of ( x_4 ), ( x_1 ), and ( x_2 ):[ x_3 = 3x_4 - 3x_2 + x_1 ]Substituting this into the sum ( x_3 + x_4 = a + 1 ):[ 3x_4 - 3x_2 + x_1 + x_4 = a + 1 ][ 4x_4 - 3x_2 + x_1 = a + 1 ]Using the sum of the roots of the first equation ( x_1 + x_2 = 1 - a ), we express ( x_1 ) as:[ x_1 = 1 - a - x_2 ]Substituting ( x_1 ) into the equation:[ 4x_4 - 3x_2 + 1 - a - x_2 = a + 1 ][ 4x_4 - 4x_2 + 1 - a = a + 1 ][ 4x_4 - 4x_2 - a = a ][ 4(x_4 - x_2) = 2a ][ x_4 - x_2 = frac{a}{2} ]Thus, the value of ( x_4 - x_2 ) is (boxed{A}).]

🔑:1. We start with the given equations: [ x^2 + ax + 2 = x ] and [ (x - a)^2 + a(x - a) + 2 = x. ]2. Simplify the first equation: [ x^2 + ax + 2 = x implies x^2 + (a-1)x + 2 = 0. ] Let the roots of this equation be (x_1) and (x_2). By Vieta's formulas, we have: [ x_1 + x_2 = -(a-1) = 1 - a ] and [ x_1 x_2 = 2. ]3. Simplify the second equation: [ (x - a)^2 + a(x - a) + 2 = x implies x^2 - 2ax + a^2 + ax - a^2 + 2 = x implies x^2 - (a+1)x + 2 = 0. ] Let the roots of this equation be (x_3) and (x_4). By Vieta's formulas, we have: [ x_3 + x_4 = a + 1 ] and [ x_3 x_4 = 2. ]4. We are given the condition: [ x_3 - x_1 = 3(x_4 - x_2). ]5. Express (x_1) and (x_3) in terms of (x_2) and (x_4): [ x_1 = 1 - a - x_2 ] and [ x_3 = a + 1 - x_4. ]6. Substitute these into the given condition: [ (a + 1 - x_4) - (1 - a - x_2) = 3(x_4 - x_2). ]7. Simplify the equation: [ a + 1 - x_4 - 1 + a + x_2 = 3(x_4 - x_2) implies 2a - x_4 + x_2 = 3(x_4 - x_2). ]8. Combine like terms: [ 2a - x_4 + x_2 = 3x_4 - 3x_2 implies 2a = 4x_4 - 4x_2 implies 4(x_4 - x_2) = 2a implies x_4 - x_2 = frac{a}{2}. ]Conclusion:[boxed{frac{a}{2}}]

🧠:[<think>Okay, so I have this system of equations to solve. Let me first write them down again to make sure I have them right:1. (left(-frac{x^{7}}{y}right)^{ln (-y)} = x^{2 ln left(x y^{2}right)})2. (y^{2} + 2 x y - 3 x^{2} + 12 x + 4 y = 0)Hmm, the first equation looks complicated with logarithms and exponents. The second one is a quadratic equation in terms of x and y. Let me start by trying to simplify the first equation. First, I notice that there's a logarithm of (-y), which implies that (-y) must be positive. So, ( -y > 0 Rightarrow y < 0 ). Therefore, y has to be negative. That's important information because it restricts the possible values of y.Now, let me look at the first equation again:(left(-frac{x^{7}}{y}right)^{ln (-y)} = x^{2 ln left(x y^{2}right)})Since (y) is negative, (-y) is positive, so the left-hand side is well-defined as long as (-frac{x^7}{y}) is positive. Since (y) is negative, the denominator is negative. Therefore, (-frac{x^7}{y} = frac{x^7}{-y}), and since (-y) is positive, the entire fraction is positive if (x^7) is positive. So (x^7 > 0), which implies that (x > 0). So both x and (-y) are positive, meaning x is positive and y is negative. Good to know.So we have constraints: (x > 0), (y < 0). Let's keep that in mind.Let me take the natural logarithm of both sides of the first equation to simplify it. Taking ln on both sides:(ln left( left(-frac{x^{7}}{y}right)^{ln (-y)} right) = ln left( x^{2 ln left(x y^{2}right)} right))Using the logarithmic identity (ln(a^b) = b ln a), this becomes:(ln (-y) cdot ln left(-frac{x^{7}}{y}right) = 2 ln left(x y^{2}right) cdot ln x)Let me simplify each side step by step.First, the left-hand side (LHS):(ln (-y) cdot ln left( -frac{x^7}{y} right))But since (y < 0), (-y = |y|), and (-frac{x^7}{y} = frac{x^7}{|y|}), because (y) is negative. So both arguments inside the logarithms are positive, as required.But let me write it more explicitly. Let me denote (z = -y), so (z > 0). Then, substituting (y = -z), the original equation becomes:(left( frac{x^7}{z} right)^{ln z} = x^{2 ln left( x (-z)^2 right)})But ( (-z)^2 = z^2 ), so inside the logarithm on the right-hand side (RHS):(x^{2 ln (x z^2)})Therefore, the equation is:(left( frac{x^7}{z} right)^{ln z} = x^{2 ln (x z^2)})But maybe substituting like this complicates things. Let me instead work with the original variables, keeping in mind that (y = -z) where (z > 0).Alternatively, perhaps expanding the logarithms directly.Starting again with the LHS:(ln (-y) cdot ln left(-frac{x^7}{y}right))Let me compute (ln left( -frac{x^7}{y} right)). Since (y) is negative, as before, this is equal to (ln left( frac{x^7}{-y} right)). So:(ln left( frac{x^7}{-y} right) = ln x^7 - ln (-y) = 7 ln x - ln (-y))Therefore, the LHS becomes:(ln (-y) cdot left(7 ln x - ln (-y)right))Similarly, the RHS:(2 ln (x y^2) cdot ln x)First, compute (ln (x y^2)). Since (y) is negative, (y^2) is positive, so (x y^2) is positive as (x > 0). Therefore:(ln (x y^2) = ln x + ln y^2 = ln x + 2 ln |y|). But since (y) is negative, (ln y^2 = 2 ln |y|). However, (ln (x y^2)) is the same as (ln x + 2 ln |y|) because (y^2 = |y|^2). But actually, if I consider (y) as a negative number, then (y^2 = (-|y|)^2 = |y|^2), so (ln y^2 = 2 ln |y|). Therefore:(ln (x y^2) = ln x + 2 ln |y|). But since (y) is negative, (|y| = -y), so:(ln (x y^2) = ln x + 2 ln (-y))Therefore, the RHS becomes:(2 (ln x + 2 ln (-y)) cdot ln x = 2 ln x (ln x + 2 ln (-y)))So putting it all together, our equation becomes:Left-Hand Side (LHS): (ln (-y) cdot (7 ln x - ln (-y)))Right-Hand Side (RHS): (2 ln x cdot (ln x + 2 ln (-y)))So the equation is:(ln (-y) cdot (7 ln x - ln (-y)) = 2 ln x cdot (ln x + 2 ln (-y)))Let me denote (a = ln x) and (b = ln (-y)), since both (x > 0) and (-y > 0), so (a) and (b) are real numbers. Then the equation becomes:(b(7a - b) = 2a(a + 2b))Let's expand both sides:Left side: (7ab - b^2)Right side: (2a^2 + 4ab)So bringing all terms to one side:(7ab - b^2 - 2a^2 - 4ab = 0)Simplify:( (7ab - 4ab) - b^2 - 2a^2 = 0 Rightarrow 3ab - b^2 - 2a^2 = 0 )Rearranged:(-2a^2 + 3ab - b^2 = 0)Multiply both sides by -1 to make it a bit more standard:(2a^2 - 3ab + b^2 = 0)Now, this is a quadratic equation in terms of a and b. Let me see if I can factor this.Looking at (2a^2 - 3ab + b^2), let's try to factor:We can treat this as quadratic in a:(2a^2 - 3ab + b^2 = 0)Let me attempt to factor:Looking for factors of the form (2a - mb)(a - nb) = 0. Let's check:Multiply out: (2a - mb)(a - nb) = 2a^2 - (2nb + m) a + m n b^2Comparing coefficients:- Coefficient of a^2: 2- Coefficient of ab: -(2n + m)- Coefficient of b^2: m nWe need:-2n - m = -3 (since coefficient of ab is -3)and m n = 1 (since coefficient of b^2 is 1)So, solving:m n = 1and 2n + m = 3From m n =1, m = 1/n.Substitute into second equation:2n + 1/n = 3Multiply both sides by n:2n^2 +1 = 3nSo:2n^2 -3n +1=0Solve quadratic equation:n = [3 ± sqrt(9 - 8)] /4 = [3 ±1]/4Thus:n = (3+1)/4=1 or n=(3-1)/4=0.5So n=1 or n=0.5.If n=1, then m=1/n=1. Then factors would be (2a -1*b)(a -1*b)= (2a -b)(a -b)Check: (2a -b)(a -b) = 2a^2 -2ab -ab + b^2=2a^2-3ab +b^2, which matches.If n=0.5, then m=1/0.5=2. Then factors would be (2a -2b)(a -0.5b). Let's check:(2a -2b)(a -0.5b) = 2a(a) +2a(-0.5b) -2b(a) -2b(-0.5b) = 2a^2 -ab -2ab +b^2=2a^2 -3ab +b^2, also correct.But since the quadratic factors into (2a -b)(a -b)=0, or equivalently (2a -2b)(a -0.5b)=0, either way. Let's take the first factorization for simplicity:(2a - b)(a - b) = 0Therefore, either:1. 2a - b = 0 ⇒ 2a = bor2. a - b = 0 ⇒ a = bRecall that (a = ln x) and (b = ln(-y)). Therefore:Case 1: 2a = b ⇒ 2 ln x = ln(-y) ⇒ ln(-y) = 2 ln x ⇒ -y = e^{2 ln x} = x^2 ⇒ y = -x^2Case 2: a = b ⇒ ln x = ln(-y) ⇒ x = -y ⇒ y = -xSo from the first equation, we have two possibilities: either y = -x^2 or y = -x.Now, let's take these two cases and substitute them into the second equation of the system:(y^{2} + 2 x y - 3 x^{2} + 12 x + 4 y = 0)Let me handle each case separately.Case 1: y = -x^2Substitute y = -x^2 into the second equation:First, compute each term:- (y^2 = (-x^2)^2 = x^4)- (2xy = 2x(-x^2) = -2x^3)- (-3x^2) remains as is- (12x) remains as is- (4y = 4(-x^2) = -4x^2)Putting all together:(x^4 - 2x^3 - 3x^2 + 12x - 4x^2 = 0)Simplify like terms:(x^4 - 2x^3 -3x^2 -4x^2 +12x = x^4 -2x^3 -7x^2 +12x =0)Factor out an x:(x(x^3 -2x^2 -7x +12) =0)So, either x=0 or (x^3 -2x^2 -7x +12 =0)But x >0 (from earlier constraints), so x=0 is rejected. Therefore, solve the cubic equation:(x^3 -2x^2 -7x +12 =0)Let me try to factor this. Possible rational roots are ±1, ±2, ±3, ±4, ±6, ±12.Testing x=1: 1 -2 -7 +12=4 ≠0x=2:8 -8 -14 +12= -2≠0x=3:27 -18 -21 +12=0. 27-18=9, 9-21=-12, -12+12=0. Yes! x=3 is a root.Therefore, factor as (x -3)(quadratic). Use polynomial division or synthetic division.Divide x^3 -2x^2 -7x +12 by (x -3):Coefficients: 1 | -2 | -7 |12Bring down 1. Multiply by 3: 3. Add to -2: 1Multiply by 3: 3. Add to -7: -4Multiply by 3: -12. Add to 12:0. Perfect.Thus, the cubic factors as (x -3)(x^2 +x -4)=0Set equal to zero: (x -3)(x^2 +x -4)=0Therefore, solutions are x=3 and roots of x^2 +x -4=0Solve x^2 +x -4=0:x = [-1 ± sqrt(1 +16)]/2 = [-1 ± sqrt(17)]/2So roots are x=( -1 + sqrt(17) )/2 ≈ ( -1 +4.123)/2≈1.5615 and x=( -1 - sqrt(17))/2≈ negative number.But x must be positive (since x >0), so the valid roots are x=3 and x=( -1 + sqrt(17))/2≈1.5615Therefore, in Case 1, possible x values are x=3 and x=( -1 + sqrt(17))/2. Let me denote sqrt(17) as approximately 4.123, so ( -1 +4.123)/2≈1.5615Now, compute y for each x:For x=3: y=-x²= -9For x=( -1 + sqrt(17))/2: y= -x². Let me compute x² first.Let me denote x = [sqrt(17) -1]/2, then x² = ([sqrt(17)-1]/2)^2 = (17 -2 sqrt(17) +1)/4 = (18 -2 sqrt(17))/4 = (9 - sqrt(17))/2Therefore, y= -x²= (-9 + sqrt(17))/2But need to check if these solutions satisfy the original equations, especially since we squared or manipulated equations, which can sometimes introduce extraneous solutions.But first, let me note that x=3, y=-9 and x=(sqrt(17)-1)/2≈1.5615, y=(-9 + sqrt(17))/2≈(-9 +4.123)/2≈-2.4385. Since y has to be negative, both solutions for y are negative, which is acceptable.Now, let's check if these solutions satisfy the original first equation.Check x=3, y=-9:First equation: (-x^7 / y)^{ln(-y)} = x^{2 ln(x y²)}Left side: (-3^7 / (-9))^{ln(9)}. Compute numerator: -3^7 = -2187, denominator: -9. So (-2187)/(-9)=243. So left side: 243^{ln 9}Right side: 3^{2 ln(3*(-9)^2)}. Compute inside the log: 3*(81)=243. So ln(243). Therefore, right side: 3^{2 ln243} = (3^{ln243})^2 = (243^{ln3})^2. Wait, maybe better to compute exponents.Wait, let me compute both sides numerically.Left side: 243^{ln9}. Compute ln9≈2.1972. So 243^2.1972≈243^2 *243^0.1972≈59049 * (243^0.1972). 243^0.1972≈ e^{0.1972 * ln243}≈ e^{0.1972 *5.493}≈e^{1.083}≈2.952. Therefore, left≈59049*2.952≈174,344.Right side: 3^{2 ln243}. Compute ln243≈5.493. So 2*5.493≈10.986. Then 3^{10.986}≈3^10 *3^0.986≈59049 *3^0.986. 3^0.986≈e^{0.986*ln3}≈e^{0.986*1.0986}≈e^{1.083}≈2.952. Therefore, right≈59049*2.952≈174,344. So both sides are equal. So x=3, y=-9 is valid.Now check x=(sqrt(17)-1)/2≈1.5615, y=(-9 + sqrt(17))/2≈(-9 +4.123)/2≈-2.4385.First equation: (-x^7 / y)^{ln(-y)} = x^{2 ln(x y²)}Compute left side:First, compute -x^7 / y. x≈1.5615, y≈-2.4385. So x^7≈1.5615^7. Let me compute step by step:1.5615^2≈2.4381.5615^3≈1.5615*2.438≈3.8081.5615^4≈1.5615*3.808≈5.9421.5615^5≈1.5615*5.942≈9.2751.5615^6≈1.5615*9.275≈14.4851.5615^7≈1.5615*14.485≈22.62So -x^7 ≈-22.62, divided by y≈-2.4385: (-22.62)/(-2.4385)=9.277≈9.277Then, ln(-y)=ln(2.4385)≈0.891Therefore, left side≈9.277^0.891≈ e^{0.891*ln9.277}≈e^{0.891*2.228}≈e^{1.985}≈7.28Right side: x^{2 ln(x y²)}. Compute x≈1.5615, y≈-2.4385. Then y²≈5.946. So x y²≈1.5615*5.946≈9.277. Then ln(x y²)=ln(9.277)≈2.228. Multiply by 2:4.456. So right side= x^{4.456}=1.5615^{4.456}. Compute:1.5615^2≈2.4381.5615^4≈(2.438)^2≈5.9421.5615^4.456≈1.5615^4 *1.5615^0.456≈5.942*(1.5615^0.456). 1.5615^0.456≈e^{0.456*ln1.5615}≈e^{0.456*0.446}≈e^{0.203}≈1.225. So total≈5.942*1.225≈7.28Thus, both sides≈7.28, so the equation holds. Therefore, this solution is also valid.So Case 1 gives two solutions: (3, -9) and approximately (1.5615, -2.4385). But let's write the exact values.x=(sqrt(17)-1)/2, y=(-9 + sqrt(17))/2. Wait, actually:Wait, earlier in Case 1, when x=( -1 + sqrt(17) )/2, then y= -x². Let me compute x²:x=(sqrt(17)-1)/2, so x²=(sqrt(17)-1)^2 /4=(17 - 2 sqrt(17) +1)/4=(18 -2 sqrt(17))/4=(9 -sqrt(17))/2Therefore, y= -x²= (-9 + sqrt(17))/2So exact form is y=(sqrt(17)-9)/2. That's approximately (4.123-9)/2≈-2.4385, which matches the approximate value.So Case 1 gives two solutions: (3, -9) and ((sqrt(17)-1)/2, (sqrt(17)-9)/2 )Now, let's check Case 2: y = -xSubstitute y = -x into the second equation:(y^2 + 2xy -3x^2 +12x +4y=0)Replace y with -x:Compute each term:- (y^2 = (-x)^2 =x^2)- (2xy =2x(-x) =-2x^2)- (-3x^2) remains- (12x) remains- (4y =4(-x)=-4x)Putting all together:(x^2 -2x^2 -3x^2 +12x -4x =0)Simplify like terms:x^2 -2x^2 -3x^2 = (1-2-3)x^2 = -4x^212x -4x=8xThus, the equation becomes:-4x^2 +8x=0 ⇒ -4x(x -2)=0Solutions: x=0 or x=2But x>0, so x=0 is rejected. Therefore, x=2, then y=-x= -2Check if (2, -2) satisfies the first equation.First equation: (-x^7 / y)^{ln(-y)} =x^{2 ln(x y²)}Compute left side: (-2^7 / (-2))^{ln(2)}. Compute numerator: -128 / (-2)=64. So left side:64^{ln2}Right side:2^{2 ln(2*(-2)^2)}. Compute inside the log:2*(4)=8. So ln8. Thus, right side:2^{2 ln8}= (2^{ln8})^2=8^{2 ln2} Hmm, wait.Wait, maybe compute both sides numerically.Left side:64^{ln2}. ln2≈0.693, so 64^0.693. Since 64=2^6, so 64^{ln2}= (2^6)^{ln2}=2^{6 ln2}=2^{ln 2^6}=2^{ln64}=64. Wait, but 2^{ln64}? Wait, perhaps another approach.Wait, 64^{ln2}=e^{ln64 * ln2}=e^{(4.1589 *0.693)}≈e^{2.883}≈17.82Right side:2^{2 ln8}. ln8≈2.079, so 2*2.079≈4.158. Therefore, 2^{4.158}=e^{4.158 ln2}≈e^{4.158*0.693}≈e^{2.883}≈17.82Therefore, both sides≈17.82, so the equation holds. Therefore, (2, -2) is a valid solution.So Case 2 gives another solution: (2, -2)Therefore, in total, the system has three solutions:1. (3, -9)2. ((sqrt(17)-1)/2, (sqrt(17)-9)/2 )3. (2, -2)But let me verify if all these solutions indeed satisfy both equations. We checked (3, -9) and (2, -2), and the approximate one. Let me verify the exact form for the second solution.Take x=(sqrt(17)-1)/2 and y=(sqrt(17)-9)/2. Let's check the second equation:y² +2xy -3x² +12x +4y=0Compute each term:First, compute x and y:x=(sqrt(17)-1)/2, y=(sqrt(17)-9)/2Compute y²:[(sqrt(17)-9)/2]^2 = (17 -18 sqrt(17) +81)/4=(98 -18 sqrt(17))/4=(49 -9 sqrt(17))/2Compute 2xy:2 * [ (sqrt(17)-1)/2 ] * [ (sqrt(17)-9)/2 ] = [ (sqrt(17)-1)(sqrt(17)-9) ] / 2Multiply numerator:sqrt(17)*sqrt(17)=17sqrt(17)*(-9) + (-1)*sqrt(17) + (-1)*(-9)= -9 sqrt(17) - sqrt(17) +9= -10 sqrt(17) +9Thus, 2xy= [17 -10 sqrt(17) +9]/2= (26 -10 sqrt(17))/2=13 -5 sqrt(17)Compute -3x²:-3*[ (sqrt(17)-1)^2 /4 ]= -3*(17 -2 sqrt(17) +1)/4= -3*(18 -2 sqrt(17))/4= (-54 +6 sqrt(17))/4= (-27 +3 sqrt(17))/2Compute 12x:12*[ (sqrt(17)-1)/2 ]=6*(sqrt(17)-1)=6 sqrt(17)-6Compute 4y:4*[ (sqrt(17)-9)/2 ]=2*(sqrt(17)-9)=2 sqrt(17)-18Now sum all terms:y² +2xy -3x² +12x +4y =(49 -9 sqrt(17))/2 + (13 -5 sqrt(17)) + (-27 +3 sqrt(17))/2 + (6 sqrt(17)-6) + (2 sqrt(17)-18)Combine all terms:First, let's convert all terms to halves to combine:= (49 -9 sqrt(17))/2 + (26 -10 sqrt(17))/2 + (-27 +3 sqrt(17))/2 + (12 sqrt(17)-12)/2 + (4 sqrt(17)-36)/2Wait, actually, let's re-express each term over 2:y²=(49 -9 sqrt(17))/22xy=13 -5 sqrt(17)= (26 -10 sqrt(17))/2-3x²=(-27 +3 sqrt(17))/212x=6 sqrt(17)-6=(12 sqrt(17)-12)/24y=2 sqrt(17)-18=(4 sqrt(17)-36)/2Therefore, sum:[ (49 -9 sqrt(17)) + (26 -10 sqrt(17)) + (-27 +3 sqrt(17)) + (12 sqrt(17)-12) + (4 sqrt(17)-36) ] / 2Compute numerator:49 +26 -27 -12 -36 = (49+26)=75; (75 -27)=48; (48 -12)=36; (36 -36)=0Now sqrt(17) terms:-9 sqrt(17) -10 sqrt(17) +3 sqrt(17) +12 sqrt(17) +4 sqrt(17)=(-9 -10 +3 +12 +4) sqrt(17)= ( (-19) +19 ) sqrt(17)=0Therefore, numerator=0 +0=0. Hence, the entire expression=0/2=0. So the equation holds. Perfect, the exact solution satisfies the second equation.Therefore, all three solutions are valid.So compiling all the solutions:1. x=3, y=-92. x=(sqrt(17)-1)/2, y=(sqrt(17)-9)/23. x=2, y=-2But the problem says "Solve the system of equations", so I need to present these solutions. However, let me check if these are all the solutions. Are there any other possibilities?Wait, when we factored the first equation, we got two cases, and in each case, we substituted into the second equation and found solutions. Since the second equation is quadratic, each case may give multiple solutions. We found three solutions in total. Since the original system is a combination of exponential and quadratic equations, it's possible to have multiple solutions. Therefore, these three are the solutions.Therefore, the solutions are:(3, -9), ((sqrt(17)-1)/2, (sqrt(17)-9)/2), and (2, -2)But let me present them in exact form. For the second solution:x=(sqrt(17)-1)/2 and y=(sqrt(17)-9)/2Alternatively, we can write them as:x= (√17 -1)/2, y= (√17 -9)/2So, boxed answers:First solution: (boxed{(3, -9)})Second solution: (boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)})Third solution: (boxed{(2, -2)})But let me check if all three are necessary. Since the problem says "Solve the system", which usually requires all solutions. So all three are valid and should be presented.Final AnswerThe solutions are (boxed{(3, -9)}), (boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}), and (boxed{(2, -2)}).</think>To solve the given system of equations:[left{begin{array}{l}left(-frac{x^{7}}{y}right)^{ln (-y)}=x^{2 ln left(x y^{2}right)} y^{2}+2 x y-3 x^{2}+12 x+4 y=0end{array}right.]we start by analyzing the first equation. Taking the natural logarithm of both sides and simplifying, we find that (y) must be negative and (x) must be positive. We then use substitutions and factorizations to reduce the equation to two cases: (y = -x^2) and (y = -x).Case 1: (y = -x^2)Substituting (y = -x^2) into the second equation, we solve the resulting polynomial equation:[x^4 - 2x^3 - 7x^2 + 12x = 0]Factoring out (x) and solving the cubic equation, we find the solutions (x = 3) and (x = frac{sqrt{17} - 1}{2}). Corresponding (y) values are (y = -9) and (y = frac{sqrt{17} - 9}{2}).Case 2: (y = -x)Substituting (y = -x) into the second equation, we solve the resulting quadratic equation:[-4x^2 + 8x = 0]This gives the solution (x = 2) with corresponding (y = -2).Verifying all solutions in the original equations confirms their validity. The solutions to the system are:[boxed{(3, -9)}, boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}, text{ and } boxed{(2, -2)}],<think>Okay, so I need to solve this system of equations. Let me first look at both equations and see how to approach them. The first equation is an exponential equation with logarithms in the exponents, which seems pretty complicated. The second equation is a quadratic in terms of y and x, but also mixed with linear terms. Hmm, this might be challenging, but let me take it step by step.Starting with the first equation:left(-frac{x^{7}}{y}right)^{ln (-y)} = x^{2 ln left(x y^{2}right)}First thing I notice is that there's a logarithm of (-y), which means that -y has to be positive. So, y must be negative. That's an important point because it restricts the possible values of y. So, y < 0. Also, inside the logarithm, we have (-y), so we must make sure that y is negative. Additionally, the term inside the first parenthesis is -x^7/y. Since y is negative, -x^7 divided by y would be positive because a negative divided by a negative is positive. So, the base of the exponent on the left side is positive, which is good because otherwise, raising a negative number to a real exponent could be problematic.Now, let me rewrite both sides using properties of exponents and logarithms to see if I can simplify the equation.Starting with the left side:left(-frac{x^{7}}{y}right)^{ln (-y)} = left(frac{x^{7}}{-y}right)^{ln (-y)}Since y is negative, -y is positive, so we can write it as:left(frac{x^{7}}{|y|}right)^{ln |y|}But maybe keeping it as is for now. Let me take the natural logarithm of both sides to bring down the exponents. That might make things easier. So, take ln of both sides:ln left( left(-frac{x^{7}}{y}right)^{ln (-y)} right) = ln left( x^{2 ln (x y^{2})} right)Using logarithm properties: ln(a^b) = b*ln(a). So left side becomes:ln (-y) cdot ln left( -frac{x^{7}}{y} right) = 2 ln (x y^{2}) cdot ln xHmm, okay. Let me simplify both sides step by step.First, the left side:ln (-y) cdot left[ ln (-x^{7}) - ln y right]Wait, hold on. Wait, inside the logarithm on the left side is -x^7 / y. Wait, no, the argument is -x^7 divided by y. Wait, original expression was -x^7 / y. But since y is negative, as we established earlier, then -x^7 / y is positive. So, actually, maybe we can write that term as positive, so maybe:Left side: ln(-y) multiplied by ln(x^7 / (-y)), since -x^7 / y = x^7 / (-y) because y is negative.Wait, let's clarify this:Original base is -x^7 / y. Since y is negative, let's denote y = -k where k > 0. Then, substitute y = -k. Let's try that substitution because it might make things clearer.Let me let y = -k, where k > 0. Then, the first equation becomes:Left side: (-x^7 / y)^{ln(-y)} = (-x^7 / (-k))^{ln(k)} = (x^7 / k)^{ln(k)}.Right side: x^{2 ln(x y^2)} = x^{2 ln(x (-k)^2)} = x^{2 ln(x k^2)}.So substituting y = -k simplifies the equation a bit. Let's write it out:Left side: (x^7 / k)^{ln k} = x^{7 ln k} / k^{ln k}Right side: x^{2 [ln x + 2 ln k]} = x^{2 ln x + 4 ln k} = x^{2 ln x} cdot x^{4 ln k} = e^{2 (ln x)^2} cdot e^{4 ln k cdot ln x}Wait, maybe another approach. Let me take logarithms of both sides again, even after substitution.Wait, maybe it's better to take logarithms again with substitution.Original equation after substitution:(x^7 / k)^{ln k} = x^{2 ln(x k^2)}Take natural logarithm of both sides:ln[(x^7 / k)^{ln k}] = ln[x^{2 ln(x k^2)}]Left side: ln k * ln(x^7 / k) = ln k * [7 ln x - ln k]Right side: 2 ln(x k^2) * ln x = 2 [ln x + 2 ln k] * ln x = 2 (ln x)^2 + 4 ln k * ln xSo, left side: 7 ln k * ln x - (ln k)^2Right side: 2 (ln x)^2 + 4 ln k * ln xSo bringing all terms to one side:7 ln k * ln x - (ln k)^2 - 2 (ln x)^2 - 4 ln k * ln x = 0Simplify:(7 ln k * ln x - 4 ln k * ln x) - 2 (ln x)^2 - (ln k)^2 = 0Which is:3 ln k * ln x - 2 (ln x)^2 - (ln k)^2 = 0Let me rearrange terms:-2 (ln x)^2 + 3 ln k * ln x - (ln k)^2 = 0This is a quadratic equation in terms of ln x. Let's denote a = ln x and b = ln k. Then, equation becomes:-2 a^2 + 3 b a - b^2 = 0Multiply both sides by -1 to make it more standard:2 a^2 - 3 b a + b^2 = 0Now, solving for a:Quadratic equation: 2a^2 - 3b a + b^2 = 0Using quadratic formula:a = [3b ± sqrt(9b^2 - 8b^2)] / 4 = [3b ± sqrt(b^2)] / 4 = [3b ± |b|] / 4But since k > 0, then b = ln k is a real number, but k could be greater than 1 or between 0 and 1. However, sqrt(b^2) = |b|.Therefore:a = [3b + |b|]/4 or a = [3b - |b|]/4Case 1: b ≥ 0 (i.e., k ≥ 1)Then |b| = b, so:a = (3b + b)/4 = 4b/4 = bOr a = (3b - b)/4 = 2b/4 = b/2Case 2: b < 0 (i.e., 0 < k < 1)Then |b| = -b, so:a = [3b - b]/4 = 2b/4 = b/2Or a = [3b + (-b)] /4 = 2b/4 = b/2Wait, wait, let's check that again.Case 2: If b < 0,First solution: [3b + |b|]/4 = [3b + (-b)] /4 = 2b/4 = b/2Second solution: [3b - |b|]/4 = [3b - (-b)] /4 = 4b/4 = bSo regardless of the sign of b, the solutions are a = b and a = b/2Therefore, the solutions are:Either a = b, which implies ln x = ln k ⇒ x = kOr a = b/2, which implies ln x = (ln k)/2 ⇒ x = sqrt(k)But remember that x must be positive because we have ln x in the original equations. Wait, in the original first equation, we have x in the exponent and in the base. Since x is raised to a power, x must be positive. Otherwise, x^7 could be negative if x is negative, but then we are raising a negative number to a real exponent, which is not defined in real numbers. So, x must be positive. Therefore, x > 0, y < 0.So, going back, x = k or x = sqrt(k)But y = -k, so k = -y. Therefore, substituting back k = -y:Case 1: x = k = -y ⇒ x = -yCase 2: x = sqrt(k) = sqrt(-y)So now we have two cases to consider. Let me note these:Case 1: x = -yCase 2: x = sqrt(-y)Now, let's substitute each case into the second equation of the system:Second equation:y^{2} + 2 x y - 3 x^{2} + 12 x + 4 y = 0So let's handle each case separately.Case 1: x = -ySubstitute x = -y into the second equation:y^2 + 2*(-y)*y - 3*(-y)^2 + 12*(-y) + 4y = 0Simplify term by term:y^2 + 2*(-y^2) - 3y^2 + (-12y) + 4yCompute each term:1. y^22. -2y^23. -3y^24. -12y5. +4yCombine like terms:For y^2 terms: 1 - 2 -3 = (1 - 5) = -4y^2For y terms: -12y +4y = -8ySo equation becomes:-4y^2 -8y = 0Factor out -4y:-4y(y + 2) = 0Set equal to zero:-4y(y + 2) = 0 ⇒ y = 0 or y = -2But earlier we established that y must be negative (y < 0), so y = 0 is rejected. Thus, y = -2.Therefore, in Case 1, y = -2, so x = -y = 2.So one solution is (x, y) = (2, -2)Now, check if this satisfies the first equation.Check the first equation with x = 2, y = -2:Left side: (- (2)^7 / (-2))^{ln(-(-2))} = (-128 / -2)^{ln(2)} = (64)^{ln 2}Right side: (2)^{2 ln(2*(-2)^2)} = 2^{2 ln(2*4)} = 2^{2 ln(8)} = (2^{ln 8})^2 = (8^{ln 2})^2 = 8^{2 ln 2} = (8^{ln 2})^2Wait, but 64^{ln 2} vs 8^{2 ln 2}Note that 64 = 8^2, so 64^{ln 2} = (8^2)^{ln 2} = 8^{2 ln 2}, which is the same as the right side. Therefore, both sides are equal. So (2, -2) is a valid solution.Case 2: x = sqrt(-y)Let me write x = sqrt(-y). Let's express y in terms of x. Let me square both sides:x^2 = -y ⇒ y = -x^2Now, substitute y = -x^2 into the second equation:y^2 + 2xy -3x^2 +12x +4y = 0Substitute y = -x^2:(-x^2)^2 + 2x*(-x^2) -3x^2 +12x +4*(-x^2) = 0Simplify each term:1. (-x^2)^2 = x^42. 2x*(-x^2) = -2x^33. -3x^24. 12x5. 4*(-x^2) = -4x^2Combine all terms:x^4 -2x^3 -3x^2 +12x -4x^2Combine like terms:x^4 -2x^3 -7x^2 +12x = 0Factor this equation:x(x^3 - 2x^2 -7x +12) = 0Set equal to zero:x = 0 or x^3 -2x^2 -7x +12 = 0But x = 0 would imply y = -0^2 = 0, but y must be negative, so x = 0 is rejected. Therefore, we need to solve the cubic equation:x^3 -2x^2 -7x +12 = 0Let me try to factor this. Let's use Rational Root Theorem. Possible rational roots are ±1, ±2, ±3, ±4, ±6, ±12.Test x=1: 1 -2 -7 +12 = 4 ≠ 0x=2: 8 -8 -14 +12 = -2 ≠0x=3: 27 -18 -21 +12 = 0. 27-18=9, 9-21=-12, -12+12=0. Yes, x=3 is a root.Therefore, we can factor (x -3) out of the cubic.Use polynomial division or synthetic division.Divide x^3 -2x^2 -7x +12 by (x -3):Using synthetic division:3 | 1 -2 -7 12 3 3 -12 1 1 -4 0So the cubic factors as (x -3)(x^2 +x -4) =0Set each factor equal to zero:x -3=0 ⇒ x=3x^2 +x -4=0 ⇒ x = [-1 ± sqrt(1 +16)]/2 = [-1 ± sqrt(17)]/2So possible x values: 3, [-1 + sqrt(17)]/2, [-1 - sqrt(17)]/2Now, since x must be positive (as established earlier), let's check the roots:x=3 is positive.For [-1 + sqrt(17)]/2: sqrt(17) is approx 4.123, so -1 +4.123 ≈3.123, divided by 2 ≈1.5615, positive.For [-1 - sqrt(17)]/2: this is negative, since sqrt(17)+1 is positive, so negative divided by 2 is negative. So reject this root.Therefore, the valid x solutions in Case 2 are x=3 and x ≈1.5615. But let's write exact values:x = 3 and x = [-1 + sqrt(17)]/2Now, since y = -x^2, let's find corresponding y:For x=3: y = -9For x=(-1 + sqrt(17))/2: y = - [ (-1 + sqrt(17))/2 ]^2Let me compute that:First, square the term:[ (-1 + sqrt(17))/2 ]^2 = [1 - 2 sqrt(17) +17 ] /4 = (18 - 2 sqrt(17))/4 = (9 - sqrt(17))/2Thus, y = - (9 - sqrt(17))/2 = (-9 + sqrt(17))/2 ≈ (-9 +4.123)/2 ≈ (-4.877)/2≈-2.438, which is negative, so acceptable.Now, need to check if these solutions satisfy the first equation.First, check x=3, y=-9.Left side of first equation: (-3^7 / (-9))^{ln(-(-9))} = (-2187 / -9)^{ln 9} = (243)^{ln9}Right side: 3^{2 ln(3*(-9)^2)} = 3^{2 ln(3*81)} = 3^{2 ln(243)} = (3^{ln243})^2But 243 = 3^5, so ln243 =5 ln3, so 3^{5 ln3} = (3^{ln3})^5 = e^{ln3 * ln3 *5} ? Wait, maybe better to compute exponents.Wait, let me compute left and right sides:Left side: 243^{ln9} = (3^5)^{ln9} = 3^{5 ln9} = 3^{5 * 2 ln3} (since 9=3^2, ln9=2ln3) = 3^{10 ln3}Right side: 3^{2 ln243} = 3^{2 *5 ln3} (since 243=3^5, ln243=5ln3) = 3^{10 ln3}Therefore, both sides equal 3^{10 ln3}, so equal. Thus, (3, -9) is valid.Now, check x = (-1 + sqrt(17))/2, y = (-9 + sqrt(17))/2This is going to be more complicated. Let me denote x = [sqrt(17) -1]/2 and y = [sqrt(17) -9]/2First, check if the first equation holds.Left side: (-x^7 / y)^{ln(-y)} = [ -x^7 / y ]^{ln(-y)}. Since y is negative, -y is positive, so ln(-y) is defined.Compute -x^7 / y:Since y = [sqrt(17) -9]/2, which is negative, so -y = [9 - sqrt(17)]/2, which is positive.Compute numerator: -x^7 = - [ (sqrt(17)-1)/2 ]^7Denominator: y = [sqrt(17)-9]/2Thus, -x^7 / y = [ - (sqrt(17)-1)^7 / 2^7 ] / [ (sqrt(17)-9)/2 ] = [ - (sqrt(17)-1)^7 / 2^7 ] * [2 / (sqrt(17)-9) ] = - (sqrt(17)-1)^7 / (2^6 * (sqrt(17)-9))This seems very messy. Maybe there's a smarter way to verify if this satisfies the first equation.Alternatively, perhaps using substitution and relationships. Remember that in Case 2, we had x = sqrt(-y), so x^2 = -y. Let me check if with this relationship, the first equation holds.Original first equation:Left side: (-x^7 / y)^{ln(-y)} = (-x^7 / y)^{ln(-y)}But since y = -x^2, substitute y:Left side: (-x^7 / (-x^2))^{ln(x^2)} = (x^7 / x^2)^{ln(x^2)} = (x^5)^{2 ln x} = x^{10 ln x}Right side: x^{2 ln(x y^2)}. Substitute y = -x^2:Right side: x^{2 ln(x*(-x^2)^2)} = x^{2 ln(x*x^4)} = x^{2 ln(x^5)} = x^{2 *5 ln x} = x^{10 ln x}Therefore, both sides are equal. Therefore, any solution in Case 2 where x = sqrt(-y) will automatically satisfy the first equation. Therefore, the solutions found in Case 2 are valid.Wait, that's an important point. If in Case 2, x = sqrt(-y), then substituting back into the first equation leads to both sides being equal, so the first equation is satisfied for any x and y in this relationship. However, we found specific solutions from the second equation, so those are the specific points where both equations are satisfied. Therefore, the solutions (3, -9) and ([sqrt(17)-1]/2, [sqrt(17)-9]/2) are valid.Therefore, the system has three solutions:1. (2, -2)2. (3, -9)3. ([sqrt(17)-1]/2, [sqrt(17)-9]/2)Wait, but wait, in Case 1, we found (2, -2), and in Case 2, x=3, y=-9 and x=(sqrt(17)-1)/2, y=(sqrt(17)-9)/2. So total three solutions. But let me verify the third solution.Wait, but the cubic equation had roots x=3, x=(−1+sqrt(17))/2, and x=(−1−sqrt(17))/2. But the last one is negative, so we rejected it. So only x=3 and x=(−1+sqrt(17))/2 are positive.Therefore, three solutions? Wait, no. In Case 1, one solution (2, -2). In Case 2, two solutions (3, -9) and ([sqrt(17)-1]/2, [sqrt(17)-9]/2). So total three solutions. But wait, let's check if ([sqrt(17)-1]/2, [sqrt(17)-9]/2) is indeed a solution.Compute y = [sqrt(17)-9]/2 ≈ (4.123 -9)/2 ≈ (-4.877)/2 ≈ -2.438, which is negative, so valid.Now, check first equation with these approximate values:x ≈ (4.123 -1)/2 ≈3.123/2≈1.5615y≈-2.438Left side: (-x^7 / y)^{ln(-y)}Compute x^7 ≈1.5615^7. Let's approximate:1.5615^2 ≈2.4381.5615^4 ≈ (2.438)^2≈5.9451.5615^6 ≈5.945 *2.438≈14.491.5615^7≈14.49 *1.5615≈22.62So -x^7 ≈-22.62, divided by y≈-2.438 gives (-22.62)/(-2.438)≈9.28ln(-y)=ln(2.438)≈0.891So left side≈9.28^0.891≈9.28^0.891≈approx 9.28^(1 -0.109)≈9.28 / 9.28^0.109. Let's compute 9.28^0.109:Take natural log: 0.109*ln9.28≈0.109*2.228≈0.243. Then exponentiate: e^0.243≈1.275Thus, 9.28 /1.275≈7.27Right side: x^{2 ln(x y^2)}Compute x≈1.5615, y≈-2.438, so y^2≈5.945x y^2≈1.5615*5.945≈9.28ln(x y^2)=ln(9.28)≈2.228Multiply by 2: 4.456Thus, x^{4.456}≈1.5615^{4.456}Compute 1.5615^2≈2.4381.5615^4≈(2.438)^2≈5.9451.5615^4.456≈5.945 *1.5615^(0.456). Compute 1.5615^0.456:Take ln:0.456*ln1.5615≈0.456*0.446≈0.203. Exponentiate: e^0.203≈1.225Thus, 5.945*1.225≈7.28Which is close to left side≈7.27. The slight discrepancy is due to approximate calculations, but they are effectively equal. Therefore, this solution is valid.Therefore, the system has three solutions: (2, -2), (3, -9), and ([sqrt(17)-1]/2, [sqrt(17)-9]/2). Wait, but let me check if ([sqrt(17)-1]/2, [sqrt(17)-9]/2) is actually a solution. Since when we derived Case 2, we substituted x = sqrt(-y) into the second equation and found these x-values. Since substituting into the first equation shows they satisfy it, as shown algebraically earlier, then they are valid.Therefore, the system has three solutions. Wait, but hold on, in the original problem statement, are we supposed to give all real solutions, or are there constraints?The problem says "Solve the system of equations," so likely all real solutions. So three solutions.Wait, but let me check again.Wait, in Case 1, we had (2, -2). In Case 2, (3, -9) and ([sqrt(17)-1]/2, [sqrt(17)-9]/2). So three solutions. Wait, but in the original substitution for Case 2, we set x = sqrt(-y), which led us to two solutions, but perhaps that's correct.Wait, maybe double-check if the cubic equation was correctly factored.Original cubic equation after substitution in Case 2: x^3 -2x^2 -7x +12=0.We found x=3 is a root. Then factoring gives (x -3)(x^2 +x -4)=0. Then solving x^2 +x -4=0 gives x=( -1 ± sqrt(17))/2. So positive roots are x=3 and x=(-1 + sqrt(17))/2. Therefore, two solutions in Case 2. So total solutions: 1 (Case 1) + 2 (Case 2) = 3 solutions.But let me check if x=3 and y=-9 are indeed solutions.Plugging into second equation:y^2 +2xy -3x^2 +12x +4y = (-9)^2 +2*3*(-9) -3*(3)^2 +12*3 +4*(-9) =81 -54 -27 +36 -36=81 -54=27; 27 -27=0; 0 +36=36; 36 -36=0. So yes, equals zero.Similarly, plugging ([sqrt(17)-1]/2, [sqrt(17)-9]/2) into the second equation. But we already know that from substitution, so they must satisfy the equation.Therefore, the system has three solutions.But wait, in the first case, substituting x=2 and y=-2 into the second equation:y² +2xy -3x² +12x +4y = (-2)^2 +2*2*(-2) -3*(2)^2 +12*2 +4*(-2) =4 -8 -12 +24 -8= (4-8)= -4; (-4-12)= -16; (-16+24)=8; (8-8)=0. So it works.Therefore, the three solutions are:1. (2, -2)2. (3, -9)3. ( [sqrt(17)-1]/2 , [sqrt(17)-9]/2 )But the problem statement doesn't specify whether to present them as exact values or decimal approximations. Since the third solution involves sqrt(17), it's better to present it in exact form.So, writing the third solution as:x = (sqrt(17) - 1)/2, y = (sqrt(17) - 9)/2Alternatively, we can write them as:x = frac{sqrt{17} -1}{2}, y = frac{sqrt{17} -9}{2}Therefore, the system of equations has three solutions: (2, -2), (3, -9), and the third one with sqrt(17). But let me check if all these solutions are distinct. Since sqrt(17) is irrational, the third solution is distinct from the others. So, three solutions in total.But wait, let me check if the third solution might coincide with one of the others. For example, if sqrt(17) ≈4.123, then (sqrt(17)-1)/2≈1.5615, which is different from x=2 and x=3. Similarly, y≈-2.438, which is different from -2 and -9. Therefore, all three solutions are distinct.Therefore, the final answer is the three ordered pairs:(2, -2), (3, -9), and (frac{sqrt{17} -1}{2}, frac{sqrt{17} -9}{2})But let me check if the problem might have only two solutions, perhaps I made a miscalculation. Let me confirm again:In Case 1: x=2, y=-2In Case 2: x=3, y=-9 and x=(sqrt(17)-1)/2, y=(sqrt(17)-9)/2Thus, three solutions. Therefore, the answer should include all three. But let me check if there is an error in the quadratic solution for the first equation.Wait, when we solved the quadratic equation 2a^2 -3ab +b^2=0, we found solutions a=b and a=b/2, leading to x=k and x=sqrt(k). But k=-y, so x=-y or x=sqrt(-y). However, when we substituted these into the second equation, we got three solutions because in Case 2, we had two solutions from the cubic. But actually, the cubic came from substituting x=sqrt(-y) into the second equation, leading to a quartic equation x^4 -2x^3 -7x^2 +12x =0, which factored as x(x^3 -2x^2 -7x +12)=0, giving x=0 (invalid) and the cubic. The cubic had three roots, but only two positive ones. So two solutions in Case 2. So total three solutions.Hence, the system has three real solutions: (2, -2), (3, -9), and ((sqrt(17)-1)/2, (sqrt(17)-9)/2). Therefore, the answer is these three ordered pairs.But let me check the problem statement again. The problem is from an exam worth 5 points. Sometimes, such problems might expect integer solutions or simpler forms, and the third solution might be considered extraneous. Wait, but when we checked, it was valid.Alternatively, maybe I made a mistake in the substitution. Let me check again.Wait, in Case 2, x = sqrt(-y), so y = -x². Substituting into the second equation:y² +2xy -3x² +12x +4y = (-x²)^2 +2x(-x²) -3x² +12x +4(-x²) =x^4 -2x³ -3x² +12x -4x²= x^4 -2x³ -7x² +12x=0, which factors to x(x³ -2x² -7x +12)=0. Then, x=0 or the cubic. Solving the cubic gave x=3 and x=(−1+sqrt(17))/2. Therefore, two solutions in Case 2. So total solutions are three. Therefore, the answer should include all three.Alternatively, maybe the problem expects only integer solutions? Let's see. (2, -2) and (3, -9) are integer solutions. The third solution is irrational. Depending on the context, maybe only integer solutions are required. But since the problem doesn't specify, we should present all solutions.Therefore, the solutions are:boxed{(2, -2)}, boxed{(3, -9)}, and boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}But since the problem says "put your final answer within boxed{}", and it's a system of equations, the standard is to present all solutions. If multiple solutions, they can be boxed separately or as a set. However, in some contexts, multiple boxed answers might be needed, but since this is a single answer field, perhaps list all solutions inside a single box.But instructions say: "put your final answer within boxed{}". So, if there are multiple solutions, we need to present them all in one box, probably separated by commas.Therefore, the final answer is:boxed{(2, -2)}, boxed{(3, -9)}, and boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}But maybe the system expects three separate boxed answers? However, given the initial instruction, it's more likely to present all solutions in one box. In LaTeX, multiple boxed answers can be written as:boxed{(2, -2)}, boxed{(3, -9)}, boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}But since the user said "put your final answer within boxed{}", which is singular, maybe they expect all solutions in one box. So, we can write:The solutions are boxed{(2, -2)}, boxed{(3, -9)}, and boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}.But given the format, maybe:boxed{(2, -2)}, boxed{(3, -9)}, and boxed{left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right)}.Alternatively, if they need to be in a single box with set notation:boxed{left{ (2, -2), (3, -9), left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right) right}}But I need to check standard conventions. In mathematics, when presenting multiple solutions to a system, they are often enclosed in set notation. Therefore, the most appropriate way is:boxed{ left( 2, -2 right) }, boxed{ left( 3, -9 right) }, and boxed{ left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right) }But given the instruction says "put your final answer within boxed{}", which implies a single box. But the system has three solutions, so maybe as a set inside a single box.Therefore, the answer is:boxed{ left{ left( 2, -2 right), left( 3, -9 right), left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right) right} }Yes, that's likely the correct format.Final AnswerThe solutions to the system are boxed{ left{ left( 2, -2 right), left( 3, -9 right), left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right) right} }.</think>To solve the given system of equations:left{begin{array}{l}left(-frac{x^{7}}{y}right)^{ln (-y)}=x^{2 ln left(x y^{2}right)} y^{2}+2 x y-3 x^{2}+12 x+4 y=0end{array}right.we start by analyzing the first equation. We note that ( y ) must be negative. By substituting ( y = -k ) (where ( k > 0 )), we simplify the first equation and find that it leads to two cases: ( x = -y ) or ( x = sqrt{-y} ).Case 1: ( x = -y )Substituting ( x = -y ) into the second equation:y^2 + 2(-y)y - 3(-y)^2 + 12(-y) + 4y = 0Simplifying, we get:-4y^2 - 8y = 0 implies y = -2Thus, ( x = 2 ), leading to the solution ( (2, -2) ).Case 2: ( x = sqrt{-y} )Substituting ( x = sqrt{-y} ) (or ( y = -x^2 )) into the second equation:(-x^2)^2 + 2x(-x^2) - 3x^2 + 12x + 4(-x^2) = 0Simplifying, we get:x^4 - 2x^3 - 7x^2 + 12x = 0Factoring, we find the roots ( x = 3 ) and ( x = frac{-1 + sqrt{17}}{2} ). Corresponding ( y ) values are ( y = -9 ) and ( y = frac{sqrt{17} - 9}{2} ).Verifying these solutions in the first equation confirms they are valid. Therefore, the solutions are:1. ( (2, -2) )2. ( (3, -9) )3. ( left( frac{sqrt{17} - 1}{2}, frac{sqrt{17} - 9}{2} right) )Thus, the solutions to the system are:boxed{ left{ left( 2, -2 right), left( 3, -9 right), left( dfrac{sqrt{17} - 1}{2}, dfrac{sqrt{17} - 9}{2} right) right} }]

🔑:Answer: (2 ;-2),(3 ;-9),left(frac{sqrt{17}-1}{2} ; frac{sqrt{17}-9}{2}right).Solution. We take the natural logarithm of the first equation in the system:ln left(-frac{x^{7}}{y}right) cdot ln (-y)=ln x^{2} cdot ln left(x y^{2}right)This equation, within the domain of valid values, is equivalent to the following:begin{aligned}& (7 ln x-ln (-y)) ln (-y)=2 ln x(ln x+2 ln (-y)) Leftrightarrow ln ^{2}(-y)-3 ln x ln (-y)+2 ln ^{2} x=0 Leftrightarrow & quad(ln (-y)-ln x)(ln (-y)-2 ln x)=0 Leftrightarrow(-y-x)left(-y-x^{2}right)=0 Leftrightarrowleft[begin{array}{l}y=-x y=-x^{2}end{array}right.end{aligned}We rewrite the second equation as y^{2}+2 y(x+2)-3 x^{2}+12 x=0 and solve it as a quadratic equation in terms of the variable y:frac{D}{4}=(x+2)^{2}-left(12 x-3 x^{2}right)=x^{2}+4 x+4-12 x+3 x^{2}=(2 x-2)^{2} Rightarrow y=-(x+2) pm(2 x-2)Thus, y=-3 x or y=x-4. There are four possible cases:a) left{begin{array}{l}y=-x, y=-3 xend{array} Leftrightarrowleft{begin{array}{l}x=0, y=0 .end{array} quadright.right. The point (0 ; 0) does not satisfy the domain of valid values.b) left{begin{array}{l}y=-x, y=x-4end{array} Leftrightarrowleft{begin{array}{l}x=2, y=-2 .end{array}right.right.c) left{begin{array}{l}y=-x^{2}, y=-3 xend{array} Leftrightarrowleft[begin{array}{l}left{begin{array}{l}x=0, y=0,end{array}right. left{begin{array}{l}x=3, y=-9 .end{array}right.end{array}right.right. The point (0 ; 0) does not satisfy the domain of valid values.d) left{begin{array}{l}y=-x^{2}, y=x-4end{array} Leftrightarrowleft{begin{array}{l}y=x-4, x^{2}+x-4=0end{array} Leftrightarrowleft{begin{array}{l}y=x-4, x=frac{-1 pm sqrt{17}}{2}end{array} Leftrightarrowleft[begin{array}{l}left{begin{array}{l}x=frac{-1+sqrt{17}}{2}, y=frac{-9+sqrt{17}}{2},end{array}right. left{begin{array}{l}x=frac{-1-sqrt{17}}{2}, y=frac{-9-sqrt{17}}{2},end{array}right.end{array}right.right.right.right.The point left(frac{-1-sqrt{17}}{2} ; frac{-9-sqrt{17}}{2}right) does not satisfy the domain of valid values.Combining the results, we get the final answer: (2 ;-2),(3 ;-9),left(frac{sqrt{17}-1}{2} ; frac{sqrt{17}-9}{2}right).

🧠:[<think>Okay, so I need to find a natural number x (positive integer) such that the integer part of its square root, denoted [√x], is equal to (2015 - x)/x. Hmm, let's break this down step by step.First, let me recall that the integer part of a number, also called the floor function, is the greatest integer less than or equal to that number. So, [√x] is the largest integer less than or equal to √x. Let's denote that integer as n. So, n = [√x]. That means n ≤ √x < n + 1. Squaring all parts, we get n² ≤ x < (n + 1)². So x is in the interval [n², (n + 1)²).Given that [√x] = (2015 - x)/x, and we set n = [√x], then we have n = (2015 - x)/x. Let's rewrite this equation:n = (2015 - x)/xMultiply both sides by x:n * x = 2015 - xBring all terms to one side:n * x + x = 2015Factor x:x(n + 1) = 2015So, x = 2015 / (n + 1)But x must be a positive integer, so (n + 1) must be a divisor of 2015. Therefore, n + 1 divides 2015. So first, let's factorize 2015 to find its divisors.2015 is an odd number. Let's check divisibility by 5 first. 2015 divided by 5 is 403. So, 2015 = 5 * 403.Now check 403: 403 divided by 13 is 31, because 13*31 = 403. So, 2015 factors into 5 * 13 * 31. Therefore, the divisors of 2015 are all the products of these primes:1, 5, 13, 31, 5*13=65, 5*31=155, 13*31=403, and 5*13*31=2015.Therefore, the positive divisors of 2015 are 1, 5, 13, 31, 65, 155, 403, 2015. Thus, the possible values for (n + 1) are these divisors, so n + 1 ∈ {1, 5, 13, 31, 65, 155, 403, 2015}, which implies that n ∈ {0, 4, 12, 30, 64, 154, 402, 2014}.But wait, n is the integer part of √x, and since x is a positive integer, √x is at least 1, so n must be at least 0. However, if n=0, then x = 2015/(0 +1) = 2015. But then [√2015] is the integer part of √2015. Let's compute √2015. Since 44² = 1936 and 45² = 2025. So √2015 is between 44 and 45, so [√2015] is 44. But if n=0, then according to the equation, n should equal 0. But 44 ≠ 0, which is a contradiction. Therefore, n=0 is invalid.So n must be in {4, 12, 30, 64, 154, 402, 2014}. Let's check each possible n:Starting with n=4: Then x=2015/(4+1)=2015/5=403. Now check if [√403] =4.Compute √400=20, so wait, wait, 20²=400, 21²=441. So √403 is between 20 and 21. Therefore, [√403]=20. But n=4, so 20≠4. So this is invalid. Hmm, something's wrong here.Wait, hold on. If n=4, then x=403. But [√403] is 20, not 4. So this does not satisfy the original equation. So this suggests that our logic is missing something.Wait, but earlier, we had that x=2015/(n+1), and that n= [√x]. Therefore, we need to have that n ≤ √x <n+1. So if x=403, then √403≈20.07, so [√403]=20. Therefore, n should be 20, but according to our previous calculation, n=4. So this is a contradiction.Therefore, the mistake is that when we set n = [√x], and derived x=2015/(n+1), but then we need to also satisfy that n ≤ √x <n+1, which translates to n² ≤x <(n+1)^2. So x must be in [n², (n+1)^2) and also x=2015/(n+1). Therefore, we need x=2015/(n+1) to be in [n², (n+1)^2). So combining these two:n² ≤ 2015/(n +1) < (n +1)^2.Therefore, we can write two inequalities:1) n² ≤ 2015/(n +1)2) 2015/(n +1) < (n +1)^2So let's solve these inequalities for n.First inequality: n²(n +1) ≤ 2015Second inequality: 2015 < (n +1)^3So, the second inequality gives (n +1)^3 > 2015. Let's compute cube roots. 12³=1728, 13³=2197. So (n+1)^3 >2015 implies n+1 ≥13, since 12³=1728 <2015 and 13³=2197>2015. Therefore, n+1 ≥13, so n ≥12.First inequality: n²(n+1) ≤2015. Let's check n=12: 12²*13=144*13=1872. 1872 ≤2015: yes.n=13: 13²*14=169*14=2366. 2366 >2015: no. So n can be at most 12. Wait, but from the second inequality, n+1 ≥13, so n≥12. Therefore, combining both, n must be exactly 12.Wait, this is conflicting. If second inequality requires n+1 ≥13, meaning n≥12, but first inequality requires n²(n+1) ≤2015. For n=12, n²(n+1)=144*13=1872 ≤2015. For n=13, 169*14=2366 >2015. So n can only be 12. Therefore, n=12.Therefore, the only possible n is 12. Let's check x=2015/(12 +1)=2015/13=155.Now, check if [√155]=12. Compute 12²=144, 13²=169. √155 is between 12 and 13, so [√155]=12. Correct. Then, check the original equation: [√155] = (2015 -155)/155.Compute (2015 -155)/155 = 1860 /155 = 12. So 12=12. Correct.Therefore, x=155 is a solution. Are there any other solutions?Wait, according to our previous analysis, the divisors of 2015 are 1,5,13,31,65,155,403,2015. Thus, n +1 could be 13, leading to n=12. Then other divisors: Let's check if for example n+1=5, which would lead to n=4. Then x=2015/5=403. But earlier, we saw that [√403]=20≠4. So invalid. Similarly, n+1=31, n=30. Then x=2015/31=65. Check [√65]=8. So 8 vs n=30. Not equal. So invalid.Similarly, n+1=65, n=64. x=2015/65=31. [√31]=5≠64. Not valid. n+1=155, n=154. x=2015/155=13. [√13]=3≠154. Not valid. n+1=403, n=402. x=2015/403=5. [√5]=2≠402. Not valid. n+1=2015, n=2014. x=2015/2015=1. [√1]=1≠2014. Not valid.So the only valid solution is x=155. Therefore, the answer is 155.But wait, let's just verify again.Given x=155, [√155]=12. Compute (2015 -155)/155 = (1860)/155 = 12. So 12=12. Correct. Therefore, 155 is indeed the solution.But why did our initial approach list n as {4,12,30,64,154,402,2014}, but only n=12 works? Because even though n +1 divides 2015, x=2015/(n+1) must also lie in [n², (n+1)^2). So even though for other n's, x is integer, x does not lie in the interval required for n to be the integer part of √x. Therefore, n=12 is the only one that satisfies both conditions.Hence, the only solution is x=155.Final AnswerThe natural number ( x ) that satisfies the given relation is boxed{155}.</think>To determine ( x in mathbb{N}^{*} ) that satisfies the relation ( [sqrt{x}] = frac{2015 - x}{x} ), we start by denoting ( n = [sqrt{x}] ). This implies ( n leq sqrt{x} < n + 1 ), or equivalently ( n^2 leq x < (n + 1)^2 ). Given the equation ( n = frac{2015 - x}{x} ), we can rewrite it as:[n cdot x = 2015 - x implies x(n + 1) = 2015 implies x = frac{2015}{n + 1}]Since ( x ) must be a positive integer, ( n + 1 ) must be a divisor of 2015. The divisors of 2015 are 1, 5, 13, 31, 65, 155, 403, and 2015. Therefore, ( n + 1 ) can be one of these divisors, leading to ( n ) being 0, 4, 12, 30, 64, 154, 402, or 2014. However, ( n ) must also satisfy ( n^2 leq x < (n + 1)^2 ).We analyze the valid ( n ) values:1. For ( n = 12 ): - ( x = frac{2015}{12 + 1} = 155 ) - Check ( [sqrt{155}] ): ( 12^2 = 144 ) and ( 13^2 = 169 ), so ( 12 leq sqrt{155} < 13 ), thus ( [sqrt{155}] = 12 ). - Verify the equation: ( frac{2015 - 155}{155} = frac{1860}{155} = 12 ), which is correct.Other potential values of ( n ) do not satisfy the interval condition ( n^2 leq x < (n + 1)^2 ). For example:- ( n = 4 ) gives ( x = 403 ), but ( [sqrt{403}] = 20 neq 4 ).- ( n = 30 ) gives ( x = 65 ), but ( [sqrt{65}] = 8 neq 30 ).Thus, the only valid solution is ( x = 155 ).[boxed{155}],<think>Okay, so I need to find a positive integer x such that the integer part of the square root of x is equal to (2015 - x) divided by x. Hmm. Let me write that down to make it clearer.The equation is: floor(√x) = (2015 - x)/x. Here, floor(a) means the greatest integer less than or equal to a. So, floor(√x) is the integer part of the square root of x.First, let's denote floor(√x) as n, where n is a positive integer. So, n ≤ √x < n + 1. Squaring all parts, we get n² ≤ x < (n + 1)². That gives me a range for x in terms of n.Now, according to the equation given, n = (2015 - x)/x. Let me rearrange this equation. Multiply both sides by x: n * x = 2015 - x. Then, bringing all terms to one side: n * x + x = 2015. Factoring out x: x(n + 1) = 2015. So, x = 2015/(n + 1). Okay, so x must be equal to 2015 divided by (n + 1). But x has to be a positive integer, so 2015 must be divisible by (n + 1). Therefore, (n + 1) is a divisor of 2015. Let me factorize 2015 to find its divisors.2015 divided by 5 is 403. Then, 403 divided by 13 is 31. So, 2015 = 5 * 13 * 31. Therefore, the positive divisors of 2015 are all combinations of these primes. The divisors are:1, 5, 13, 31, 5*13=65, 5*31=155, 13*31=403, and 5*13*31=2015.So, the divisors are 1, 5, 13, 31, 65, 155, 403, 2015. Therefore, (n + 1) must be one of these divisors, which means n + 1 is in that list, so n is one less: 0, 4, 12, 30, 64, 154, 402, 2014. But since n is the floor of √x and x is a positive integer, n must also be a positive integer. Therefore, n can't be 0. So possible n values are 4, 12, 30, 64, 154, 402, 2014. Hmm. Wait, but n is floor(√x), so n must satisfy n² ≤ x < (n+1)². But we also have x = 2015/(n + 1). So, substituting x into the inequality:n² ≤ 2015/(n + 1) < (n + 1)².So, let's check for each possible n whether this inequality holds.First, let's list possible n values:4, 12, 30, 64, 154, 402, 2014.Let's check each n:Starting with n=4:First, calculate x = 2015/(4 + 1) = 2015/5 = 403.Check if n² ≤ x < (n + 1)².n² = 16. 16 ≤ 403 < 25? Wait, 403 is way larger than 25. So 403 is not less than 25. That's not possible. So n=4 doesn't work. Hmm. So, contradiction here. Maybe I made a mistake. Wait, n is 4, so x is 403. Then floor(√x) should be 4? Let's check √403. √400 is 20, so √403 is approximately 20.07. Therefore, floor(√403) is 20, not 4. That's a problem. So, my mistake here is that n is supposed to be floor(√x), but when I compute x as 2015/(n + 1), the resulting x may not satisfy floor(√x) = n. Therefore, I need to check for each n, compute x, then check if floor(√x) is indeed n.So, my previous approach is correct, but I need to verify after computing x whether floor(√x) is n. Let's proceed again with n=4:x=2015/(4 +1)=403. √403≈20.07, floor is 20≠4. So invalid.n=12:x=2015/(12 +1)=2015/13=155.Check floor(√155). √144=12, √169=13, so √155≈12.45. Floor is 12. So n=12 matches floor(√155)=12. So this is a possible solution.Wait, so x=155. Let me check if floor(√155)=12. Yes. Then check the original equation: floor(√155)=12. (2015 -155)/155=1860/155=12. So 12=12. Correct. So x=155 is a solution. Let's check other n's.n=30:x=2015/(30 +1)=2015/31=65.Check floor(√65). √64=8, √81=9, so floor(√65)=8. But n=30, which is not equal to 8. So invalid.n=64:x=2015/(64 +1)=2015/65=31.Check floor(√31). √25=5, √36=6. So floor(√31)=5. But n=64≠5. Invalid.n=154:x=2015/(154 +1)=2015/155=13.Check floor(√13). √9=3, √16=4. Floor is 3. n=154≠3. Invalid.n=402:x=2015/(402 +1)=2015/403=5.Floor(√5)=2. n=402≠2. Invalid.n=2014:x=2015/(2014 +1)=2015/2015=1.Floor(√1)=1. Here, n=2014≠1. So invalid.Therefore, only n=12 gives a valid solution x=155. Let me confirm once more:x=155, floor(√155)=12, and (2015 -155)/155=(1860)/155=12. So yes, 12=12. Correct.But wait, let me check if there are other n's that I might have missed. For example, perhaps n is not in the list derived from the divisors. But according to the equation x(n +1)=2015, x must be a divisor of 2015, so x must be one of the divisors:1,5,13,31,65,155,403,2015. But then, x=155 is the only one where floor(√x) equals n=(2015 -x)/x.Wait, let's check all x in the list of divisors of 2015:x=1: floor(√1)=1. (2015 -1)/1=2014≠1.x=5: floor(√5)=2. (2015 -5)/5=2010/5=402≠2.x=13: floor(√13)=3. (2015 -13)/13=2002/13=154≠3.x=31: floor(√31)=5. (2015 -31)/31=1984/31≈64.0, so 1984 divided by 31: 31*64=1984. So 64≠5.x=65: floor(√65)=8. (2015 -65)/65=1950/65=30≠8.x=155: floor(√155)=12. (2015 -155)/155=1860/155=12. Correct.x=403: floor(√403)=20. (2015 -403)/403=1612/403≈4.0. 403*4=1612. So 4≠20.x=2015: floor(√2015)=44 (since 44²=1936, 45²=2025). (2015 -2015)/2015=0/2015=0≠44.Thus, only x=155 satisfies the equation. So the answer is 155.But let me just think again if there could be another x not in the list of divisors. Wait, the equation x(n +1)=2015 implies that x is a divisor of 2015, because n +1 must be an integer (since n is integer). Therefore, x must divide 2015. Therefore, all possible x's are the divisors of 2015, which we checked, and only x=155 works.Therefore, the solution is x=155.Final AnswerThe positive integer ( x ) that satisfies the given relation is boxed{155}.</think>To determine ( x in mathbb{N}^{*} ) that satisfies the relation ( lfloor sqrt{x} rfloor = frac{2015 - x}{x} ), we start by letting ( lfloor sqrt{x} rfloor = n ), where ( n ) is a positive integer. This implies ( n leq sqrt{x} < n + 1 ), or equivalently ( n^2 leq x < (n + 1)^2 ).Given the equation ( n = frac{2015 - x}{x} ), we rearrange it to ( x(n + 1) = 2015 ). Therefore, ( x = frac{2015}{n + 1} ). Since ( x ) must be a positive integer, ( n + 1 ) must be a divisor of 2015. The divisors of 2015 are 1, 5, 13, 31, 65, 155, 403, and 2015. Thus, ( n + 1 ) can be any of these divisors, leading to possible values of ( n ) as 0, 4, 12, 30, 64, 154, 402, or 2014. However, ( n ) must be a positive integer, so we discard 0.We then check each possible ( n ) to see if ( x = frac{2015}{n + 1} ) satisfies ( n leq sqrt{x} < n + 1 ):- For ( n = 4 ), ( x = frac{2015}{5} = 403 ). Here, ( lfloor sqrt{403} rfloor = 20 neq 4 ).- For ( n = 12 ), ( x = frac{2015}{13} = 155 ). Here, ( lfloor sqrt{155} rfloor = 12 ), which matches ( n ).- For ( n = 30 ), ( x = frac{2015}{31} = 65 ). Here, ( lfloor sqrt{65} rfloor = 8 neq 30 ).- For ( n = 64 ), ( x = frac{2015}{65} = 31 ). Here, ( lfloor sqrt{31} rfloor = 5 neq 64 ).- For ( n = 154 ), ( x = frac{2015}{155} = 13 ). Here, ( lfloor sqrt{13} rfloor = 3 neq 154 ).- For ( n = 402 ), ( x = frac{2015}{403} = 5 ). Here, ( lfloor sqrt{5} rfloor = 2 neq 402 ).- For ( n = 2014 ), ( x = frac{2015}{2015} = 1 ). Here, ( lfloor sqrt{1} rfloor = 1 neq 2014 ).Only ( n = 12 ) satisfies the condition with ( x = 155 ). Verifying, ( lfloor sqrt{155} rfloor = 12 ) and ( frac{2015 - 155}{155} = 12 ), confirming the solution.Thus, the positive integer ( x ) that satisfies the given relation is (boxed{155}).]

🔑:1. [sqrt{x}]=frac{2015}{x}-1 so frac{2015}{x} in mathbb{N}^{*}, x in mathbb{N}^{*} it follows that x is a divisor of 2015 2 pSince 2015=5 cdot 13 cdot 31. By checking, it is found that among the divisors, x=5 cdot 31=155 satisfies the condition. Therefore, x=155 .5 p

🧠:[<think>Okay, so I need to find the perimeter of quadrilateral GCHF in a regular hexagon with side length 2. Let me start by visualizing the hexagon. Since it's regular, all sides are equal, and all internal angles are 120 degrees. Let me label the hexagon vertices as A, B, C, D, E, F in order. Points G and H are midpoints of AB and DE respectively. So, G is halfway between A and B, and H is halfway between D and E. The quadrilateral in question is GCHF. So the sides I need to find are GC, CH, HF, and FG. Once I have those lengths, I can add them up to get the perimeter.First, maybe I should draw the hexagon and mark all these points. But since I can't draw here, I need to do it mentally. Let me assign coordinates to the hexagon vertices. That might help. In a regular hexagon, it's often easiest to place it centered at the origin with one vertex at (r, 0), where r is the distance from the center to a vertex. But since the side length is 2, I need to relate that to the radius.Wait, in a regular hexagon, the side length is equal to the radius. So if each side is length 2, then the distance from the center to any vertex is also 2. That simplifies things. So let's set up a coordinate system with the center of the hexagon at (0, 0). Let me assign coordinates to each vertex.Starting with vertex A at (2, 0). Then moving 60 degrees around the circle for each subsequent vertex. The coordinates can be calculated using trigonometry. Each vertex will be separated by 60 degrees.So:- Vertex A: (2, 0)- Vertex B: (2 cos 60°, 2 sin 60°) = (2*(1/2), 2*(√3/2)) = (1, √3)- Vertex C: (2 cos 120°, 2 sin 120°) = (2*(-1/2), 2*(√3/2)) = (-1, √3)- Vertex D: (2 cos 180°, 2 sin 180°) = (-2, 0)- Vertex E: (2 cos 240°, 2 sin 240°) = (2*(-1/2), 2*(-√3/2)) = (-1, -√3)- Vertex F: (2 cos 300°, 2 sin 300°) = (2*(1/2), 2*(-√3/2)) = (1, -√3)Let me verify these coordinates. Each step, moving 60 degrees, so angles from 0° up. Yes, that looks right. Now, midpoints G and H.G is the midpoint of AB. Let's find coordinates of G. A is (2, 0) and B is (1, √3). The midpoint is the average of coordinates:G_x = (2 + 1)/2 = 1.5G_y = (0 + √3)/2 = √3/2So G is (1.5, √3/2)Similarly, H is the midpoint of DE. D is (-2, 0) and E is (-1, -√3). Midpoint H:H_x = (-2 + (-1))/2 = -1.5H_y = (0 + (-√3))/2 = -√3/2So H is (-1.5, -√3/2)Now, the quadrilateral is GCHF. Let's note the coordinates of each point in GCHF:- G: (1.5, √3/2)- C: (-1, √3)- H: (-1.5, -√3/2)- F: (1, -√3)- Back to G? Wait, the quadrilateral is GCHF, so the points are G connected to C, then to H, then to F, then back to G. So the sides are GC, CH, HF, FG.Wait, let's check that. The problem says "the perimeter of GCHF", which should be the sum of the lengths of GC, CH, HF, and FG. Wait, but the last side is from F back to G? Because GCHF is a quadrilateral. So it's G to C, C to H, H to F, and F to G. So sides: GC, CH, HF, FG. Wait, but in the problem statement, the options include 4√3, 8, 4√5, 4√7, 12. So maybe the perimeter is a combination of sides each of which is sqrt(something), adding up to one of these. Let's compute each side.First, GC. From G(1.5, √3/2) to C(-1, √3). Let's compute the distance between these two points.Distance formula: sqrt[(x2 - x1)^2 + (y2 - y1)^2]Compute x difference: (-1 - 1.5) = -2.5, which is -5/2.y difference: (√3 - √3/2) = √3/2.So distance GC is sqrt[(-5/2)^2 + (√3/2)^2] = sqrt[(25/4) + (3/4)] = sqrt[28/4] = sqrt[7]. So sqrt(7).Okay, so GC is sqrt(7). Next, CH. From C(-1, √3) to H(-1.5, -√3/2). Let's compute that.x difference: (-1.5 - (-1)) = (-0.5) = -1/2y difference: (-√3/2 - √3) = (-√3/2 - 2√3/2) = (-3√3/2)Distance CH is sqrt[(-1/2)^2 + (-3√3/2)^2] = sqrt[(1/4) + (27/4)] = sqrt[28/4] = sqrt[7]. So another sqrt(7).Hmm, same as GC. Interesting. Then HF: from H(-1.5, -√3/2) to F(1, -√3).x difference: 1 - (-1.5) = 2.5 = 5/2y difference: (-√3 - (-√3/2)) = (-√3/2)Distance HF is sqrt[(5/2)^2 + (-√3/2)^2] = sqrt[25/4 + 3/4] = sqrt[28/4] = sqrt[7]. Another sqrt(7). Finally, FG: from F(1, -√3) to G(1.5, √3/2). Let's compute this.x difference: 1.5 - 1 = 0.5 = 1/2y difference: (√3/2 - (-√3)) = (√3/2 + √3) = (3√3/2)Distance FG is sqrt[(1/2)^2 + (3√3/2)^2] = sqrt[1/4 + 27/4] = sqrt[28/4] = sqrt[7]. So all four sides of quadrilateral GCHF are sqrt(7). Therefore, the perimeter is 4*sqrt(7), which is option D. Wait, but let me verify once again. Because if all four sides are equal, that would make GCHF a rhombus. Is that possible? Let me check coordinates again.GC: sqrt[(-5/2)^2 + (sqrt3/2)^2] = sqrt[25/4 + 3/4] = sqrt[28/4] = sqrt(7). Correct.CH: sqrt[(-1/2)^2 + (-3sqrt3/2)^2] = sqrt[1/4 + 27/4] = sqrt[28/4] = sqrt(7). Correct.HF: sqrt[(5/2)^2 + (-sqrt3/2)^2] = same as GC. Correct.FG: sqrt[(1/2)^2 + (3sqrt3/2)^2] = same as CH. Correct.So all four sides are indeed sqrt(7). Therefore, the perimeter is 4*sqrt(7), which is option D. So the answer should be D.But let me just think again if there's any possibility of miscalculating the coordinates. Let me confirm coordinates of all points.Vertex A: (2, 0). Correct.Vertex B: 60 degrees from A. So (2 cos 60, 2 sin 60) = (1, sqrt3). Correct.C: 120 degrees, which is ( -1, sqrt3). Correct.D: 180 degrees, (-2, 0). Correct.E: 240 degrees, which is (-1, -sqrt3). Correct.F: 300 degrees, which is (1, -sqrt3). Correct.Midpoint G between A(2,0) and B(1, sqrt3): ((2+1)/2, (0 + sqrt3)/2) = (1.5, sqrt3/2). Correct.Midpoint H between D(-2,0) and E(-1, -sqrt3): ((-2 + (-1))/2, (0 + (-sqrt3))/2) = (-1.5, -sqrt3/2). Correct.So coordinates are all correct. Then the distances between G and C, etc., as calculated. So each distance is sqrt(7), so perimeter is 4sqrt7. So answer D.Alternatively, maybe there's a symmetry here. Since the hexagon is regular, perhaps the quadrilateral is symmetric in some way. But given that the midpoints are on different sides, maybe not rotational symmetry, but each side of the quadrilateral spans similar distances. But regardless, according to coordinates, all sides are sqrt7. So unless I made an arithmetic mistake, that's correct.Wait, just to check again the calculation for FG: from F(1, -sqrt3) to G(1.5, sqrt3/2). The y difference is sqrt3/2 - (-sqrt3) = sqrt3/2 + sqrt3 = (1/2 + 1)sqrt3 = (3/2)sqrt3. So squared, (3/2 sqrt3)^2 = (9/4)*3 = 27/4. The x difference is 0.5, squared is 0.25. So total distance sqrt(27/4 + 1/4) = sqrt(28/4) = sqrt7. Correct. So yes, all sides are sqrt7. So perimeter is 4sqrt7. So the answer is D.Final Answerboxed{D}</think>To find the perimeter of quadrilateral (GCHF) in a regular hexagon (ABCDEF) with side length 2, we start by assigning coordinates to the vertices of the hexagon. The coordinates are calculated as follows:- (A(2, 0))- (B(1, sqrt{3}))- (C(-1, sqrt{3}))- (D(-2, 0))- (E(-1, -sqrt{3}))- (F(1, -sqrt{3}))Next, we find the coordinates of the midpoints (G) and (H):- (G) is the midpoint of (overline{AB}): (left(frac{2+1}{2}, frac{0+sqrt{3}}{2}right) = left(1.5, frac{sqrt{3}}{2}right))- (H) is the midpoint of (overline{DE}): (left(frac{-2 + (-1)}{2}, frac{0 + (-sqrt{3})}{2}right) = left(-1.5, -frac{sqrt{3}}{2}right))We then calculate the distances between each pair of points forming the quadrilateral (GCHF):1. Distance (GC): [ sqrt{left(-1 - 1.5right)^2 + left(sqrt{3} - frac{sqrt{3}}{2}right)^2} = sqrt{left(-frac{5}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2} = sqrt{frac{25}{4} + frac{3}{4}} = sqrt{7} ]2. Distance (CH): [ sqrt{left(-1.5 - (-1)right)^2 + left(-frac{sqrt{3}}{2} - sqrt{3}right)^2} = sqrt{left(-frac{1}{2}right)^2 + left(-frac{3sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{27}{4}} = sqrt{7} ]3. Distance (HF): [ sqrt{left(1 - (-1.5)right)^2 + left(-sqrt{3} - left(-frac{sqrt{3}}{2}right)right)^2} = sqrt{left(frac{5}{2}right)^2 + left(-frac{sqrt{3}}{2}right)^2} = sqrt{frac{25}{4} + frac{3}{4}} = sqrt{7} ]4. Distance (FG): [ sqrt{left(1.5 - 1right)^2 + left(frac{sqrt{3}}{2} - (-sqrt{3})right)^2} = sqrt{left(frac{1}{2}right)^2 + left(frac{3sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{27}{4}} = sqrt{7} ]All four sides of quadrilateral (GCHF) are (sqrt{7}). Therefore, the perimeter is (4sqrt{7}).[boxed{D}],<think>Okay, let me try to figure out the perimeter of quadrilateral GCHF in the regular hexagon ABCDEF. The hexagon has side length 2, and G is the midpoint of AB, while H is the midpoint of DE. The answer choices are given, so maybe after working through the problem, I can check which one it matches.First, since it's a regular hexagon, all sides are equal, and all internal angles are 120 degrees. Regular hexagons can be divided into six equilateral triangles, each with side length equal to the hexagon's side length. So, maybe I can use coordinates or some geometric properties to find the lengths of the sides of quadrilateral GCHF.Let me try to sketch the hexagon first to visualize where points G and H are. Let's label the hexagon vertices clockwise as A, B, C, D, E, F, with A at the rightmost point. So, moving around the hexagon, each vertex is 60 degrees apart. Since it's regular, the center of the hexagon is equidistant from all vertices.Now, G is the midpoint of AB. Since AB is a side of length 2, G is 1 unit away from both A and B. Similarly, H is the midpoint of DE, so it's 1 unit away from D and E.Quadrilateral GCHF connects points G, C, H, and F. So, the perimeter will be the sum of the lengths of GC, CH, HF, and FG.I need to find each of these four segments' lengths. Let me try to break it down step by step.First, perhaps placing the hexagon on a coordinate system would help. Let me place point A at (2, 0) because the side length is 2, and the distance from the center to any vertex (the radius) is also 2. Wait, actually, in a regular hexagon, the side length is equal to the radius. So, if the side length is 2, then the distance from the center to any vertex is 2. Let me confirm that.In a regular hexagon, the radius (distance from center to a vertex) is equal to the side length. So yes, if each side is 2, then the radius is also 2. Therefore, placing the hexagon centered at the origin, with vertex A at (2, 0). Then, the other vertices can be found by rotating around the center by 60 degrees each time.So, coordinates of the vertices:- Point A: (2, 0)- Point B: (2*cos(60°), 2*sin(60°)) = (2*(1/2), 2*(√3/2)) = (1, √3)- Point C: (-1, √3) [Wait, hold on, adding 60° each time. Wait, starting from A at 0°, then B at 60°, C at 120°, D at 180°, E at 240°, F at 300°, back to A.]Wait, let's calculate coordinates properly using angles. Each vertex is at angle θ = 0°, 60°, 120°, 180°, 240°, 300°, etc.So, in coordinates:- A: (2 cos 0°, 2 sin 0°) = (2, 0)- B: (2 cos 60°, 2 sin 60°) = (2*(1/2), 2*(√3/2)) = (1, √3)- C: (2 cos 120°, 2 sin 120°) = (2*(-1/2), 2*(√3/2)) = (-1, √3)- D: (2 cos 180°, 2 sin 180°) = (-2, 0)- E: (2 cos 240°, 2 sin 240°) = (2*(-1/2), 2*(-√3/2)) = (-1, -√3)- F: (2 cos 300°, 2 sin 300°) = (2*(1/2), 2*(-√3/2)) = (1, -√3)Okay, that seems right. So, coordinates:A (2, 0)B (1, √3)C (-1, √3)D (-2, 0)E (-1, -√3)F (1, -√3)Now, midpoints G and H.G is the midpoint of AB. So, coordinates of G: average of coordinates of A and B.A is (2, 0), B is (1, √3). So midpoint G is ((2 + 1)/2, (0 + √3)/2) = (1.5, √3/2)Similarly, H is the midpoint of DE. D is (-2, 0), E is (-1, -√3). Midpoint H is ((-2 + (-1))/2, (0 + (-√3))/2) = (-1.5, -√3/2)Now, quadrilateral GCHF has vertices at G (1.5, √3/2), C (-1, √3), H (-1.5, -√3/2), F (1, -√3). Wait, is F at (1, -√3)? Yes, as per above.So, to find the perimeter, need to compute the lengths of GC, CH, HF, and FG.Let me compute each segment:1. GC: from G (1.5, √3/2) to C (-1, √3)Distance formula: sqrt[(x2 - x1)^2 + (y2 - y1)^2]Compute x difference: (-1 - 1.5) = -2.5y difference: (√3 - √3/2) = √3/2So, length GC = sqrt[(-2.5)^2 + (√3/2)^2] = sqrt[(6.25) + (3/4)] = sqrt[6.25 + 0.75] = sqrt[7] ≈ 2.6458Wait, 6.25 + 0.75 is 7? Yes, because 6 + 0.25 + 0.75 = 7. So sqrt(7). So GC is sqrt(7).2. CH: from C (-1, √3) to H (-1.5, -√3/2)x difference: (-1.5 - (-1)) = -0.5y difference: (-√3/2 - √3) = (-√3/2 - 2√3/2) = (-3√3)/2So, length CH = sqrt[(-0.5)^2 + (-3√3/2)^2] = sqrt[(0.25) + (27/4)]Convert to quarters: 0.25 = 1/4, 27/4 is 27/4.So total inside sqrt: 1/4 + 27/4 = 28/4 = 7Therefore, length CH = sqrt(7)3. HF: from H (-1.5, -√3/2) to F (1, -√3)x difference: 1 - (-1.5) = 2.5y difference: -√3 - (-√3/2) = -√3/2So, length HF = sqrt[(2.5)^2 + (-√3/2)^2] = sqrt[6.25 + 3/4] = sqrt[7] again. Same as before.4. FG: from F (1, -√3) to G (1.5, √3/2)x difference: 1.5 - 1 = 0.5y difference: √3/2 - (-√3) = √3/2 + √3 = (√3 + 2√3)/2 = 3√3/2So, length FG = sqrt[(0.5)^2 + (3√3/2)^2] = sqrt[0.25 + (27/4)]Again, convert 0.25 to 1/4, so total inside sqrt: 1/4 + 27/4 = 28/4 = 7. Therefore, sqrt(7).So all four sides of quadrilateral GCHF are sqrt(7), so the perimeter is 4*sqrt(7), which is option D.Wait, but let me double-check these calculations to make sure I didn't make a mistake.Starting with GC:G is (1.5, √3/2), C is (-1, √3)x difference: -1 - 1.5 = -2.5y difference: √3 - √3/2 = √3/2Then squared terms: (-2.5)^2 = 6.25, (√3/2)^2 = 3/46.25 + 0.75 = 7. sqrt(7). Correct.Then CH:C is (-1, √3), H is (-1.5, -√3/2)x difference: -1.5 - (-1) = -0.5y difference: -√3/2 - √3 = -3√3/2Squares: (-0.5)^2 = 0.25, (-3√3/2)^2 = (9*3)/4 = 27/40.25 + 27/4 = 1/4 + 27/4 = 28/4 = 7. sqrt(7). Correct.HF:H (-1.5, -√3/2) to F (1, -√3)x difference: 1 - (-1.5) = 2.5y difference: -√3 - (-√3/2) = -√3 + √3/2 = -√3/2Squares: (2.5)^2 = 6.25, (-√3/2)^2 = 3/4. 6.25 + 0.75 = 7. sqrt(7). Correct.FG:F (1, -√3) to G (1.5, √3/2)x difference: 1.5 - 1 = 0.5y difference: √3/2 - (-√3) = √3/2 + √3 = 3√3/2Squares: 0.5^2 = 0.25, (3√3/2)^2 = 27/40.25 + 27/4 = 1/4 + 27/4 = 28/4 =7. sqrt(7). Correct.So all four sides are sqrt(7). Therefore, perimeter is 4*sqrt(7), which is answer choice D. So the answer should be D.Wait, but just to be thorough, maybe I should check if the quadrilateral is indeed a four-sided figure with all sides equal? That would make it a rhombus, but depending on the angles, it could be a square or something else. But in this case, given the coordinates, it's a rhombus with all sides sqrt(7), so perimeter is 4 times that. So 4*sqrt7. So the answer is D.Final Answerboxed{D}</think>To find the perimeter of quadrilateral (GCHF) in a regular hexagon (ABCDEF) with side length 2, we start by determining the coordinates of the vertices and midpoints.1. Coordinates of Hexagon Vertices: - (A(2, 0)) - (B(1, sqrt{3})) - (C(-1, sqrt{3})) - (D(-2, 0)) - (E(-1, -sqrt{3})) - (F(1, -sqrt{3}))2. Midpoints Calculation: - (G) is the midpoint of (overline{AB}): [ Gleft(frac{2 + 1}{2}, frac{0 + sqrt{3}}{2}right) = left(1.5, frac{sqrt{3}}{2}right) ] - (H) is the midpoint of (overline{DE}): [ Hleft(frac{-2 + (-1)}{2}, frac{0 + (-sqrt{3})}{2}right) = left(-1.5, -frac{sqrt{3}}{2}right) ]3. Distance Calculations: - Distance (GC): [ sqrt{(-1 - 1.5)^2 + left(sqrt{3} - frac{sqrt{3}}{2}right)^2} = sqrt{(-2.5)^2 + left(frac{sqrt{3}}{2}right)^2} = sqrt{6.25 + 0.75} = sqrt{7} ] - Distance (CH): [ sqrt{(-1.5 - (-1))^2 + left(-frac{sqrt{3}}{2} - sqrt{3}right)^2} = sqrt{(-0.5)^2 + left(-frac{3sqrt{3}}{2}right)^2} = sqrt{0.25 + 6.75} = sqrt{7} ] - Distance (HF): [ sqrt{(1 - (-1.5))^2 + left(-sqrt{3} - left(-frac{sqrt{3}}{2}right)right)^2} = sqrt{(2.5)^2 + left(-frac{sqrt{3}}{2}right)^2} = sqrt{6.25 + 0.75} = sqrt{7} ] - Distance (FG): [ sqrt{(1.5 - 1)^2 + left(frac{sqrt{3}}{2} - (-sqrt{3})right)^2} = sqrt{(0.5)^2 + left(frac{3sqrt{3}}{2}right)^2} = sqrt{0.25 + 6.75} = sqrt{7} ]Each side of quadrilateral (GCHF) is (sqrt{7}), so the perimeter is (4sqrt{7}).Thus, the final answer is (boxed{D}).]

🔑:Let the center of the hexagon be O. triangle AOB, triangle BOC, triangle COD, triangle DOE, triangle EOF, and triangle FOA are all equilateral triangles with side length 2. Thus, CO = 2, and GO = sqrt{AO^2 - AG^2} = sqrt{3}. By symmetry, angle COG = 90^{circ}. Thus, by the Pythagorean theorem, CG = sqrt{2^2 + sqrt{3}^2} = sqrt{7}. Because CO = OF and GO = OH, CG = HC = FH = GF = sqrt{7}. Thus, the solution to our problem is sqrt{7} + sqrt{7} + sqrt{7} + sqrt{7} = boxed{textbf{(D)} 4sqrt7}.~mathboy100

🧠:[<think>Okay, so I need to find the minimum value of the function f(x) = 27x - x³ on the interval [-3, 3]. Hmm, let's see. I remember that to find the extrema (which are maximums or minimums) of a function on a closed interval, we can use the Extreme Value Theorem. It states that a continuous function on a closed interval will have both a maximum and a minimum value. Since f(x) is a polynomial, it's continuous everywhere, so definitely on [-3, 3]. That means there must be a minimum value here somewhere.The theorem also says that these extrema can occur either at critical points inside the interval or at the endpoints. Critical points are where the derivative is zero or undefined. Since the function is a polynomial, its derivative will exist everywhere, so critical points only occur where the derivative is zero. So I need to find the derivative of f(x), set it equal to zero, solve for x, and then check those x-values along with the endpoints to see which gives the smallest f(x).Let me start by finding the derivative. The function is f(x) = 27x - x³. The derivative of 27x is 27, and the derivative of -x³ is -3x². So f'(x) = 27 - 3x². Now, set the derivative equal to zero to find critical points:27 - 3x² = 0Let me solve for x. Subtract 27 from both sides:-3x² = -27Divide both sides by -3:x² = 9Take the square root of both sides:x = ±3Wait, but the interval we're looking at is [-3, 3]. So the critical points are at x = 3 and x = -3. But these are also the endpoints of the interval. Hmm, so does that mean there are no critical points inside the interval? Because 3 and -3 are endpoints, not inside the interval. So if I remember correctly, critical points inside the interval would be points where the derivative is zero and x is in (-3, 3). But since x = 3 and x = -3 are endpoints, solving f'(x) = 0 gives us x values that are exactly at the endpoints. So does that mean there are no critical points inside the interval? That seems a bit strange, but let me check my math again.Derivative f'(x) = 27 - 3x². Setting equal to zero: 27 - 3x² = 0 → x² = 9 → x = ±3. Yeah, that's correct. So indeed, the critical points are at x = 3 and x = -3, which are the endpoints. Therefore, in the interval [-3, 3], the only critical points (where derivative is zero) are at the endpoints.But wait, usually, endpoints are considered separately. The critical points inside the interval are where the derivative is zero and x is strictly between -3 and 3. But in this case, the only solutions are x = ±3, which are endpoints, so there are no critical points inside the interval. That's interesting. So in this case, the function's critical points coincide with the endpoints.Therefore, according to the Extreme Value Theorem, the minimum and maximum must occur at the endpoints since there are no critical points inside the interval. But wait, let me think again. If there are no critical points inside the interval, then all extrema must be at the endpoints. So to find the minimum, we just need to evaluate the function at x = -3 and x = 3 and see which is smaller.Wait, but maybe I made a mistake here. Let me visualize the function f(x) = 27x - x³. If I think about the graph of this function, it's a cubic function. The leading term is -x³, so as x approaches positive infinity, the function goes to negative infinity, and as x approaches negative infinity, it goes to positive infinity. But on the interval [-3, 3], let's see.At x = -3: f(-3) = 27*(-3) - (-3)^3 = -81 - (-27) = -81 + 27 = -54.At x = 3: f(3) = 27*3 - 3³ = 81 - 27 = 54.So at the endpoints, the function takes values -54 and 54. But since we are looking for the minimum value, which is the smallest value, that would be -54 at x = -3. However, wait a second. If there are no critical points inside the interval, then the endpoints are the only candidates. But I should double-check if perhaps I missed something here.Is there a chance that the function could dip lower somewhere between -3 and 3? Let's take a few sample points. For example, at x = 0: f(0) = 0 - 0 = 0. That's higher than -54. At x = 1: f(1) = 27*1 - 1 = 27 -1 = 26. Still higher. At x = -1: f(-1) = 27*(-1) - (-1)^3 = -27 - (-1) = -27 +1 = -26. Which is higher than -54. At x = -2: f(-2) = 27*(-2) - (-2)^3 = -54 - (-8) = -54 +8 = -46. Still higher than -54. Similarly, at x = 2: f(2) = 54 -8 = 46. So between -3 and 3, the function starts at -54 when x = -3, goes up to 0 at x=0, then peaks at 54 when x=3. Wait, so actually, the function is increasing from x = -3 to x = 3? But that can't be, because the derivative at x = 0 is f'(0) = 27 - 0 = 27, which is positive. So the function is increasing at x = 0, but the derivative is f'(x) = 27 -3x². So as x approaches 3 from the left, the derivative would be 27 -3*(something less than 3)^2. For example, at x = 2, f'(2) = 27 -3*4 = 27 -12 = 15. Still positive. At x approaching 3 from the left, say x=2.9: f'(2.9) = 27 -3*(8.41) = 27 -25.23 = 1.77. Still positive. At x=3, derivative is zero. Similarly, moving left from x=3, the derivative is positive until x=3. So the function is increasing all the way from x=-3 to x=3? Wait, but at x=-3, the derivative is f'(-3) = 27 - 3*(9) = 27 -27 = 0. Then, as x increases from -3 to 3, the derivative starts at 0, then becomes positive, and remains positive until x=3 where it becomes zero again. Wait, is that possible? Let me check at x=-3: f'(x)=0, then moving towards x=0, the derivative f'(x) = 27 -3x². For example, at x=-2: f'(-2) =27 -3*(4)=27-12=15, which is positive. At x=-1: 27-3*(1)=24, positive. At x=0:27. So actually, the derivative is positive throughout the interval (-3,3). Wait, so that would mean the function is increasing on the entire interval except at the endpoints where the derivative is zero. So from x=-3 to x=3, the function is increasing. Therefore, the minimum value would be at the left endpoint x=-3, which is -54, and the maximum at the right endpoint x=3, which is 54. Therefore, the minimum value is -54.But let me confirm this. Let's plug in x=-3 into the original function: f(-3) =27*(-3) - (-3)^3 = -81 - (-27) = -81 +27= -54. Correct. And since the function is increasing all the way from -3 to 3, there are no other critical points in between. Therefore, the minimum is indeed at x=-3 with value -54.But just to be thorough, let's check if there are any points where the derivative is undefined. The derivative is f'(x)=27-3x², which is a polynomial, so it's defined for all real numbers. Therefore, there are no critical points due to undefined derivatives. The only critical points are where the derivative is zero, which are x=3 and x=-3, which are the endpoints.So yes, that seems conclusive. The function is increasing on the interval [-3, 3], so the minimum is at the left end, x=-3, and the maximum at the right end, x=3. Therefore, the minimum value is -54.Final AnswerThe minimum value of the function on the interval is boxed{-54}.</think>To find the minimum value of the function ( f(x) = 27x - x^3 ) on the interval ([-3, 3]), we use the Extreme Value Theorem, which states that a continuous function on a closed interval will have both a maximum and a minimum value. First, we find the critical points by taking the derivative of the function and setting it to zero:[f'(x) = 27 - 3x^2]Setting the derivative equal to zero:[27 - 3x^2 = 0 implies x^2 = 9 implies x = pm 3]The critical points ( x = 3 ) and ( x = -3 ) are the endpoints of the interval ([-3, 3]). Next, we evaluate the function at the endpoints:- At ( x = -3 ): [ f(-3) = 27(-3) - (-3)^3 = -81 + 27 = -54 ]- At ( x = 3 ): [ f(3) = 27(3) - 3^3 = 81 - 27 = 54 ]Since the function is increasing on the interval ([-3, 3]) (as the derivative ( f'(x) = 27 - 3x^2 ) is positive for all ( x ) in ((-3, 3))), the minimum value occurs at the left endpoint ( x = -3 ).Thus, the minimum value of the function on the interval is (boxed{-54}).,<think>Okay, so I need to find the minimum value of the function f(x) = 27x - x³ on the interval [-3, 3]. Hmm, let's see. I remember that to find the minimum or maximum of a function on a closed interval, you have to check both the critical points inside the interval and the endpoints. Right? Because the extreme values can occur either where the derivative is zero or undefined, or at the endpoints. First, maybe I should find the derivative of f(x). Let's do that. The derivative of 27x is 27, and the derivative of -x³ is -3x². So f'(x) = 27 - 3x². Got that. Now, critical points are where the derivative is zero or undefined. Since this is a polynomial, it's defined everywhere, so I just need to solve for f'(x) = 0. So setting 27 - 3x² = 0. Let's solve for x. Subtract 27 from both sides: -3x² = -27. Then divide both sides by -3: x² = 9. Taking the square root of both sides, x = ±3. Wait, so the critical points are at x = 3 and x = -3. But hold on, the interval given is [-3, 3]. So these critical points are actually at the endpoints of the interval. That's interesting. So normally, we check the critical points inside the interval and the endpoints. But in this case, the critical points are exactly the endpoints. So does that mean that the only critical points are the endpoints? If so, then to find the minimum value, we just need to evaluate the function at the endpoints and see which one is smaller? Hmm, but wait, maybe I should double-check my derivative. Let me confirm. f(x) = 27x - x³. The derivative of 27x is 27, and the derivative of -x³ is indeed -3x², so f'(x) = 27 - 3x². That's correct. So solving 27 - 3x² = 0, yeah, x² = 9, so x = ±3. So those are the critical points. So in the interval [-3, 3], these critical points coincide with the endpoints. Therefore, the only points we need to check are x = -3 and x = 3? But wait, maybe I need to check for other critical points. But according to the derivative, there are no other solutions. So in this case, the function's critical points are at the endpoints. Therefore, the minimum value of the function on the interval must occur at one of these endpoints. But let me verify this. Let's evaluate f(x) at x = -3 and x = 3. First, f(-3) = 27*(-3) - (-3)³. Calculating that: 27*(-3) is -81. Then (-3)³ is -27, so subtracting that is - (-27) which is +27. So f(-3) = -81 + 27 = -54. Then f(3) = 27*(3) - (3)³. 27*3 is 81. (3)³ is 27. So 81 - 27 = 54. So f(3) = 54. Wait, so at x = -3, the function is -54, and at x = 3, it's 54. So if those are the only critical points, then on the interval [-3, 3], the function's minimum is -54 at x = -3, and the maximum is 54 at x = 3. But hold on, before concluding, maybe I should check some other points in the interval to make sure there's no lower value. For example, let's take x = 0. f(0) = 27*0 - 0³ = 0. That's higher than -54. What about x = -2? Let's compute f(-2): 27*(-2) - (-2)^3. 27*(-2) is -54. (-2)^3 is -8, so subtracting that gives - (-8) = +8. So f(-2) = -54 + 8 = -46, which is higher than -54. How about x = -1? f(-1) = 27*(-1) - (-1)^3 = -27 - (-1) = -27 +1 = -26. Still higher than -54. What about x = -3? We already checked that. How about x = -4? Wait, no, x can't be -4 because the interval is up to -3. So maybe just confirm that as we move from -3 towards 0, the function increases. Similarly, from 0 to 3, the function first increases then decreases? Wait, let's actually analyze the behavior of the function. The function f(x) = 27x - x³. Let's see, for x in [-3, 3], the derivative is 27 - 3x². So when x is in (-3, 3), the derivative is 27 - 3x². Let's see when the derivative is positive or negative. Set f'(x) = 27 - 3x². So when is this positive? When 27 > 3x² => 9 > x² => x² < 9 => -3 < x < 3. Wait, that's the entire interval except the endpoints. But the derivative at the endpoints is zero. Wait, so for x in (-3, 3), the derivative is positive? Wait, hold on. Let me compute f'(0). f'(0) = 27 - 0 = 27 > 0. So at x = 0, the function is increasing. But if x is approaching 3 from the left, say x = 2.9, then f'(2.9) = 27 - 3*(2.9)^2. Let's compute that: 2.9 squared is 8.41, 3*8.41 is 25.23. 27 - 25.23 = 1.77. Still positive. So at x approaching 3 from the left, derivative is still positive. Similarly, approaching -3 from the right, x = -2.9. Then f'(-2.9) = 27 - 3*(8.41) = same as before, 1.77. So positive. So the derivative is positive throughout the open interval (-3, 3). That means the function is increasing on the entire interval. Wait, but if the function is increasing on the entire interval from -3 to 3, then the minimum value would be at the left endpoint x = -3, and the maximum at the right endpoint x = 3. But that seems counterintuitive because the function is a cubic. Normally, a cubic function has inflection points and changes direction. But in this case, maybe due to the coefficient, on the interval [-3, 3], it's only increasing. Let's check the graph. Alternatively, think about the endpoints. Since the derivative is positive everywhere in between, the function is increasing, so yes, starting at x = -3, f(-3) = -54, and increasing all the way to x = 3, f(3) = 54. Therefore, the minimum is at x = -3, and maximum at x = 3. But wait, in that case, the answer would be -54. But let me make sure. Let me check another point, like x = -3. We already did. What about x = -2.5? f(-2.5) = 27*(-2.5) - (-2.5)^3. Calculating this: 27*(-2.5) = -67.5. (-2.5)^3 = -15.625. So subtracting that: -67.5 - (-15.625) = -67.5 +15.625 = -51.875. That's higher than -54. So as we move from -3 to -2.5, the function increases from -54 to -51.875. Similarly, moving towards 0, it keeps increasing. At x=0, it's 0. Then continues increasing up to x=3 where it's 54. So yes, the function is strictly increasing on the interval [-3, 3], so the minimum is at x=-3 and maximum at x=3. Therefore, the minimum value is -54. But wait, just to be thorough, maybe I should check if the function indeed doesn't have a local minimum somewhere else. But according to the derivative, the critical points are only at x = ±3, and the derivative is positive in between, so the function is increasing throughout the interval. Therefore, there are no local minima or maxima inside the interval. So the extremum values are only at the endpoints. Therefore, the minimum value is -54, achieved at x = -3. But just to make sure, let's consider the second derivative test. The second derivative of f(x) is f''(x) = derivative of f'(x) = derivative of 27 - 3x² = -6x. At critical points x = -3 and x = 3. At x = -3, f''(-3) = -6*(-3) = 18 > 0, which would indicate a local minimum. But x=-3 is an endpoint, so the second derivative test here might not be directly applicable for endpoints. Similarly, at x =3, f''(3) = -6*3 = -18 <0, indicating a local maximum, but again, x=3 is an endpoint. But since at x=-3, the second derivative is positive, that suggests that if we extended the function beyond x=-3, there would be a local minimum there. However, since we're confined to the interval [-3,3], the point x=-3 is the left endpoint, and since the function is increasing throughout the interval, the minimum is at x=-3. So yes, I think that's the answer. Therefore, the minimum value is -54, achieved at x = -3.Final AnswerThe minimum value of the function in the interval is boxed{-54}.</think>To find the minimum value of the function ( f(x) = 27x - x^3 ) on the interval ([-3, 3]), we follow these steps:1. Find the derivative of the function: [ f'(x) = 27 - 3x^2 ]2. Determine the critical points by setting the derivative equal to zero: [ 27 - 3x^2 = 0 implies x^2 = 9 implies x = pm 3 ] The critical points ( x = 3 ) and ( x = -3 ) are the endpoints of the interval ([-3, 3]).3. Evaluate the function at the endpoints: - At ( x = -3 ): [ f(-3) = 27(-3) - (-3)^3 = -81 + 27 = -54 ] - At ( x = 3 ): [ f(3) = 27(3) - 3^3 = 81 - 27 = 54 ]4. Analyze the behavior of the function: - The derivative ( f'(x) = 27 - 3x^2 ) is positive for all ( x ) in the open interval ((-3, 3)), indicating that the function is increasing throughout this interval.5. Conclusion: - Since the function is increasing on ([-3, 3]), the minimum value occurs at the left endpoint ( x = -3 ).Thus, the minimum value of the function in the interval is (boxed{-54}).]

🔑:Given that xin[-3,3],We have f'(x)=27-3x^{2}=-3(x+3)(x-3)geq0,Thus, the function f(x) is monotonically increasing in the interval [-3,3].Therefore, when x=-3, the function f(x) attains its minimum value, f(-3)=-3times27-(-3)^{3}=boxed{-54}.

🧠:[<think>Okay, so I need to find the values of a and b for the function f(x) = (a ln x)/(x + 1) + b/x. The tangent line at the point (1, f(1)) is given by x + 2y - 3 = 0. Hmm, let me recall how to approach this.First, I remember that the equation of the tangent line at a particular point requires two things: the value of the function at that point, and the derivative of the function at that point. So, I need to find f(1) and f'(1), then set up equations based on the tangent line equation.Let me start by finding f(1). Plugging x = 1 into the function:f(1) = (a ln 1)/(1 + 1) + b/1.But ln 1 is 0, so the first term disappears. Therefore, f(1) = b. That's straightforward.Now, the tangent line passes through the point (1, f(1)) which is (1, b). The tangent line equation given is x + 2y - 3 = 0. Let me rearrange that to slope-intercept form (y = mx + c) to find the slope and the y-intercept.Starting with x + 2y - 3 = 0,Subtract x and add 3: 2y = -x + 3,Divide by 2: y = (-1/2)x + 3/2.So, the slope of the tangent line (m) is -1/2, and when x = 1, y should be (-1/2)(1) + 3/2 = (-1/2 + 3/2) = 1. Therefore, the point (1, 1) lies on the tangent line, but since the tangent line touches the curve at (1, f(1)), that point must also be (1, 1). Therefore, f(1) = 1, which means b = 1. Wait, that seems like one value found already. So, b is 1?Hold on, let's verify. If f(1) = b, and the tangent line at x=1 has y=1 there, then yes, because when x=1, the point on the tangent line is (1,1), which is the same as the point on the curve. Therefore, f(1) must equal 1, so b=1. That's one variable found.Now, moving on to the derivative. The slope of the tangent line is -1/2, so f'(1) must be equal to -1/2. Let's compute f'(x).The function f(x) is (a ln x)/(x + 1) + b/x. Since we already found that b=1, I can substitute that in. So f(x) = (a ln x)/(x + 1) + 1/x.Now, compute the derivative f'(x). Let's do term by term.First term: (a ln x)/(x + 1). This is a quotient, so I need to use the quotient rule. The derivative of u/v is (u'v - uv')/v^2.Let u = a ln x, so u' = a*(1/x).v = x + 1, so v' = 1.Therefore, the derivative of the first term is [ (a/x)(x + 1) - a ln x * 1 ] / (x + 1)^2.Simplify numerator: (a(x + 1)/x - a ln x ) / (x + 1)^2.Second term: derivative of 1/x is -1/x^2.Therefore, putting it all together:f'(x) = [ (a(x + 1)/x - a ln x ) / (x + 1)^2 ] - 1/x^2.Now, evaluate f'(1):First, plug in x=1 into the first term's numerator:a(1 + 1)/1 - a ln 1 = a*2 - 0 = 2a.The denominator is (1 + 1)^2 = 4.So, first term becomes 2a / 4 = a/2.Second term: -1/(1)^2 = -1.Therefore, f'(1) = (a/2) - 1.But we know that the slope of the tangent line at x=1 is -1/2. Therefore:(a/2) - 1 = -1/2.Let's solve for a:Multiply both sides by 2: a - 2 = -1.Therefore, a = 1.So, a is 1 and b is 1?Wait, let me check if that's correct.Let me verify the steps again.First, f(1) = b. Since the point (1, f(1)) is on the tangent line, substituting x=1 into the tangent line equation:1 + 2f(1) - 3 = 0 => 2f(1) - 2 = 0 => f(1)=1, so b=1. That's correct.Then, f'(1) = slope of the tangent line, which is -1/2.Computed f'(x):First term derivative: [ (a/x)(x + 1) - a ln x ] / (x +1)^2. At x=1:(a/1)(1 +1) - a ln 1 = 2a - 0 = 2a. Divided by (2)^2=4, so 2a/4 = a/2.Second term derivative: -1/x^2. At x=1, that's -1.Thus, total f'(1)= a/2 -1 = -1/2.Solving a/2 -1 = -1/2:a/2 = -1/2 +1 = 1/2 => a=1. Correct.Therefore, a=1 and b=1.Wait, but let me check if plugging these back into the original function gives the correct tangent line.So, f(x)= (1 * ln x)/(x +1) + 1/x.At x=1, f(1)=0 +1=1, which matches.Compute f'(x):First term derivative: [ (1/x)(x +1) - ln x ]/(x +1)^2 -1/x^2.Simplify numerator of first term:( (x +1)/x - ln x ) = (1 + 1/x - ln x )Therefore, f'(x) = [1 + 1/x - ln x ]/(x +1)^2 -1/x^2.Wait, but when x=1:[1 +1/1 -0]/(4) -1/1 = (2)/4 -1 = 0.5 -1 = -0.5, which is -1/2. Correct.So, all steps check out. Therefore, a=1 and b=1.But wait, let me make sure there's no miscalculation here.Yes, seems right. The key steps were evaluating f(1) to find b, then computing the derivative at 1 to find a. All the algebra checks. So I think that's correct.Final AnswerThe values of ( a ) and ( b ) are (boxed{1}) and (boxed{1}), respectively.Final Answer( a = boxed{1} ) and ( b = boxed{1} )</think>Given the function ( f(x) = frac{a ln x}{x+1} + frac{b}{x} ), we need to find the values of ( a ) and ( b ) such that the tangent line to the curve ( y = f(x) ) at the point ( (1, f(1)) ) is given by the equation ( x + 2y - 3 = 0 ).First, we determine the value of ( f(1) ):[f(1) = frac{a ln 1}{1 + 1} + frac{b}{1} = 0 + b = b]Since the point ( (1, f(1)) ) lies on the tangent line, substituting ( x = 1 ) and ( y = f(1) ) into the tangent line equation ( x + 2y - 3 = 0 ) gives:[1 + 2f(1) - 3 = 0 implies 2f(1) - 2 = 0 implies f(1) = 1]Thus, ( b = 1 ).Next, we find the derivative ( f'(x) ). The function ( f(x) ) is:[f(x) = frac{a ln x}{x + 1} + frac{1}{x}]Using the quotient rule for the first term ( frac{a ln x}{x + 1} ):[frac{d}{dx} left( frac{a ln x}{x + 1} right) = frac{a cdot frac{1}{x} cdot (x + 1) - a ln x cdot 1}{(x + 1)^2} = frac{a left( frac{x + 1}{x} - ln x right)}{(x + 1)^2}]Simplifying the numerator:[frac{a left( frac{x + 1}{x} - ln x right)}{(x + 1)^2} = frac{a left( 1 + frac{1}{x} - ln x right)}{(x + 1)^2}]The derivative of the second term ( frac{1}{x} ) is:[frac{d}{dx} left( frac{1}{x} right) = -frac{1}{x^2}]Combining these, the derivative ( f'(x) ) is:[f'(x) = frac{a left( 1 + frac{1}{x} - ln x right)}{(x + 1)^2} - frac{1}{x^2}]Evaluating ( f'(1) ):[f'(1) = frac{a left( 1 + 1 - 0 right)}{(1 + 1)^2} - frac{1}{1^2} = frac{2a}{4} - 1 = frac{a}{2} - 1]The slope of the tangent line at ( x = 1 ) is ( -frac{1}{2} ). Therefore:[frac{a}{2} - 1 = -frac{1}{2}]Solving for ( a ):[frac{a}{2} = frac{1}{2} implies a = 1]Thus, the values of ( a ) and ( b ) are (boxed{1}) and (boxed{1}), respectively.,<think>Okay, let me try to solve this problem step by step. So, we have the function f(x) = (a ln x)/(x + 1) + b/x, and the tangent line at the point (1, f(1)) is given by x + 2y - 3 = 0. We need to find the values of a and b.First, I remember that to find the equation of the tangent line at a particular point, we need two things: the value of the function at that point (which gives us a point on the line) and the derivative of the function at that point (which gives us the slope of the tangent line). The equation of the tangent line can then be written using the point-slope form.Given that the tangent line is x + 2y - 3 = 0, I should probably convert that into slope-intercept form (y = mx + c) to easily identify the slope and the y-intercept. Let me rearrange the equation:Starting with x + 2y - 3 = 0,Subtract x and add 3 to both sides: 2y = -x + 3,Divide both sides by 2: y = (-1/2)x + 3/2.So, the slope (m) of the tangent line is -1/2, and the y-intercept is 3/2.Now, since the tangent line passes through the point (1, f(1)), this point should satisfy the equation of the tangent line. Therefore, substituting x = 1 into the tangent line equation will give me the corresponding y-value, which is f(1). Let me check that:Plug x = 1 into y = (-1/2)x + 3/2:y = (-1/2)(1) + 3/2 = -1/2 + 3/2 = 2/2 = 1.So, f(1) should equal 1. Let's compute f(1) using the given function:f(x) = (a ln x)/(x + 1) + b/x.Plugging x = 1:ln(1) is 0, so the first term becomes (a*0)/(1 + 1) = 0. The second term is b/1 = b. Therefore, f(1) = 0 + b = b. But we just found that f(1) = 1, so this means b = 1.Alright, so now we know that b is 1. That's one variable found. Now, we need to find a. For that, we need the derivative of f(x) evaluated at x = 1, which should equal the slope of the tangent line, which is -1/2.Let me compute the derivative f'(x). The function f(x) is (a ln x)/(x + 1) + b/x. Since we already found that b = 1, we can substitute that in: f(x) = (a ln x)/(x + 1) + 1/x. Let's compute the derivative term by term.First, the derivative of (a ln x)/(x + 1). Let me denote this as u/v, where u = a ln x and v = x + 1. The derivative of u/v is (u'v - uv') / v².Compute u':u = a ln x, so u' = a*(1/x) = a/x.v = x + 1, so v' = 1.Therefore, the derivative of the first term is [ (a/x)(x + 1) - a ln x * 1 ] / (x + 1)².Simplify numerator:(a/x)(x + 1) = a*(x + 1)/x = a*(1 + 1/x) = a + a/x.Then subtract a ln x: So numerator becomes a + a/x - a ln x.Therefore, the derivative of the first term is [a + a/x - a ln x] / (x + 1)².Now, the second term of f(x) is 1/x, so its derivative is -1/x².Therefore, putting it all together, the derivative f'(x) is [a + a/x - a ln x]/(x + 1)² - 1/x².Now, we need to evaluate this derivative at x = 1.Let's compute each part step by step.First, compute the numerator of the first term at x = 1:a + a/1 - a ln 1 = a + a - a*0 = 2a.The denominator (x + 1)² at x = 1 is (1 + 1)² = 4.So, the first term evaluates to 2a / 4 = a/2.The second term of the derivative is -1/x². At x = 1, this is -1/1² = -1.Therefore, f'(1) = (a/2) - 1.But we know that the slope of the tangent line at x = 1 is -1/2. So,f'(1) = -1/2.Setting up the equation:a/2 - 1 = -1/2.Solving for a:Add 1 to both sides: a/2 = -1/2 + 1 = 1/2.Multiply both sides by 2: a = 1.So, a is 1 and b is 1. Let me verify if this makes sense.Let me check the function with a = 1 and b = 1:f(x) = (1 * ln x)/(x + 1) + 1/x.At x = 1, f(1) = 0 + 1 = 1, which matches the point (1,1). Then, the derivative at x = 1 is (1/2) - 1 = -1/2, which matches the slope. So, plugging back into the tangent line equation:Using point-slope form: y - 1 = -1/2 (x - 1).Multiply both sides by 2: 2y - 2 = -x + 1.Rearranged: x + 2y - 3 = 0, which is exactly the given tangent line equation. So, everything checks out.Therefore, the values are a = 1 and b = 1.Final AnswerThe values of ( a ) and ( b ) are (boxed{a = 1}) and (boxed{b = 1}).</think>Given the function ( f(x) = frac{a ln x}{x + 1} + frac{b}{x} ), we need to find the values of ( a ) and ( b ) such that the equation of the tangent line to the curve ( y = f(x) ) at the point ( (1, f(1)) ) is ( x + 2y - 3 = 0 ).First, we convert the given tangent line equation ( x + 2y - 3 = 0 ) into slope-intercept form:[2y = -x + 3 implies y = -frac{1}{2}x + frac{3}{2}]This gives us the slope ( -frac{1}{2} ) and the y-intercept ( frac{3}{2} ).Next, we determine ( f(1) ). Substituting ( x = 1 ) into the function ( f(x) ):[f(1) = frac{a ln 1}{1 + 1} + frac{b}{1} = 0 + b = b]Since the point ( (1, f(1)) ) lies on the tangent line, substituting ( x = 1 ) into the tangent line equation gives ( y = 1 ). Therefore, ( f(1) = 1 implies b = 1 ).Next, we compute the derivative ( f'(x) ) to find the slope of the tangent line at ( x = 1 ). The function is:[f(x) = frac{a ln x}{x + 1} + frac{1}{x}]Using the quotient rule for the first term ( frac{a ln x}{x + 1} ):[frac{d}{dx} left( frac{a ln x}{x + 1} right) = frac{a cdot frac{1}{x} cdot (x + 1) - a ln x cdot 1}{(x + 1)^2} = frac{a left( frac{x + 1}{x} - ln x right)}{(x + 1)^2} = frac{a left( 1 + frac{1}{x} - ln x right)}{(x + 1)^2}]The derivative of the second term ( frac{1}{x} ) is:[frac{d}{dx} left( frac{1}{x} right) = -frac{1}{x^2}]Combining these, the derivative ( f'(x) ) is:[f'(x) = frac{a left( 1 + frac{1}{x} - ln x right)}{(x + 1)^2} - frac{1}{x^2}]Evaluating ( f'(1) ):[f'(1) = frac{a left( 1 + 1 - 0 right)}{(1 + 1)^2} - frac{1}{1^2} = frac{2a}{4} - 1 = frac{a}{2} - 1]Given the slope of the tangent line is ( -frac{1}{2} ), we set up the equation:[frac{a}{2} - 1 = -frac{1}{2}]Solving for ( a ):[frac{a}{2} = frac{1}{2} implies a = 1]Thus, the values of ( a ) and ( b ) are (boxed{a = 1}) and (boxed{b = 1}).]

🔑:First, let's find the derivative of the function f(x):f'(x)= frac{a(frac{x+1}{x}-ln x)}{(x+1)^{2}} - frac{b}{x^{2}}.The slope of the line x+2y-3=0 is -frac{1}{2}. Since the line passes through the point (1,1), we have:f(1)=1,and the slope of the tangent line at x=1 is equal to the derivative of f at x=1:f'(1)=-frac{1}{2}.Now let's find f(1) and f'(1):f(1) = frac{aln 1}{1+1} + frac{b}{1} = 0 + b = b = 1,so b=1.Next,f'(1) = frac{a(frac{1+1}{1}-ln 1)}{(1+1)^{2}} - frac{1}{1^{2}} = frac{a}{2} - 1.Since f'(1)=-frac{1}{2}, we have:frac{a}{2} - 1 = -frac{1}{2},which impliesa = 1.So the values of a and b are:boxed{a = 1, quad b = 1}.