answer：Okay, so I have this function f(x) = 3^x + 9^x, and it's defined on the interval [t, t + 1]. The problem says that the maximum value of f(x) on this interval is 12, and I need to find the minimum value of f(x). Hmm, let me think about how to approach this.First, I remember that functions like 3^x and 9^x are exponential functions, which are always increasing. So, their sum should also be increasing, right? Wait, but 9^x is actually (3^2)^x = 3^(2x), so it's a faster-growing exponential function. So, f(x) is the sum of two increasing functions, which means f(x) itself is increasing. If that's the case, then on the interval [t, t + 1], the function f(x) should attain its maximum at x = t + 1 and its minimum at x = t.But the problem says that the maximum value is 12. So, if f(x) is increasing, then f(t + 1) = 12. Then, the minimum value would be f(t). So, I need to find t such that f(t + 1) = 12, and then compute f(t).Let me write that down:f(t + 1) = 3^(t + 1) + 9^(t + 1) = 12.I can rewrite 9^(t + 1) as (3^2)^(t + 1) = 3^(2(t + 1)) = 3^(2t + 2).So, f(t + 1) = 3^(t + 1) + 3^(2t + 2) = 12.Let me set y = 3^t. Then, 3^(t + 1) = 3 * 3^t = 3y, and 3^(2t + 2) = (3^t)^2 * 3^2 = y^2 * 9.So, substituting back, f(t + 1) becomes:3y + 9y^2 = 12.So, 9y^2 + 3y - 12 = 0.This is a quadratic equation in terms of y. Let me write it as:9y^2 + 3y - 12 = 0.I can simplify this equation by dividing all terms by 3:3y^2 + y - 4 = 0.Now, let's solve for y using the quadratic formula. The quadratic is 3y^2 + y - 4 = 0, so a = 3, b = 1, c = -4.The discriminant D = b^2 - 4ac = (1)^2 - 4*3*(-4) = 1 + 48 = 49.So, y = [-b ± sqrt(D)] / (2a) = [-1 ± 7] / 6.So, two solutions:y = (-1 + 7)/6 = 6/6 = 1,y = (-1 - 7)/6 = -8/6 = -4/3.But y = 3^t, which is always positive, so y cannot be negative. Therefore, y = 1.So, 3^t = 1. Since 3^0 = 1, so t = 0.Therefore, t = 0. So, the interval is [0, 1].Now, since f(x) is increasing, the minimum value is at x = 0.Compute f(0):f(0) = 3^0 + 9^0 = 1 + 1 = 2.So, the minimum value is 2.Wait, let me double-check to make sure I didn't make any mistakes.First, f(t + 1) = 12. With t = 0, f(1) = 3^1 + 9^1 = 3 + 9 = 12. That's correct.And f(0) = 1 + 1 = 2, which is the minimum. Since f(x) is increasing, yes, that makes sense.Alternatively, maybe I should check if f(x) is indeed increasing on the interval. Let's compute its derivative.f(x) = 3^x + 9^x.The derivative f’(x) = ln(3) * 3^x + ln(9) * 9^x.Since ln(3) and ln(9) are positive, and 3^x and 9^x are positive for all x, f’(x) is always positive. Therefore, f(x) is strictly increasing on any interval, so the maximum is at the right endpoint and the minimum at the left endpoint.Therefore, my reasoning seems solid. So, the minimum value is 2.**Final Answer**The minimum value of ( f(x) ) is boxed{2}.

answer：Okay, so I have this arithmetic sequence problem here. Let me try to figure it out step by step. Hmm, the problem says that in an arithmetic sequence {a_n}, the sum of the first n terms is S_n. We're given that a_7 + a_9 = 16 and S_7 = 7. We need to find the value of a_{12}.Alright, let's recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a_1 and the common difference is d, then the nth term is given by a_n = a_1 + (n - 1)d. Got that.Also, the sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * (2a_1 + (n - 1)d) or sometimes written as S_n = n*(a_1 + a_n)/2. Both formulas are equivalent, right?So, we have two pieces of information here: a_7 + a_9 = 16 and S_7 = 7. Let's write down what these mean in terms of a_1 and d.First, let's express a_7 and a_9 using the formula for the nth term.a_7 = a_1 + (7 - 1)d = a_1 + 6da_9 = a_1 + (9 - 1)d = a_1 + 8dSo, a_7 + a_9 = (a_1 + 6d) + (a_1 + 8d) = 2a_1 + 14dAnd we know that this sum is 16. So, equation one is:2a_1 + 14d = 16Let me write that down:Equation 1: 2a_1 + 14d = 16Now, let's look at the second piece of information: S_7 = 7.Using the sum formula, S_7 = 7/2 * (2a_1 + (7 - 1)d) = 7/2 * (2a_1 + 6d)So, S_7 = (7/2)*(2a_1 + 6d) = 7*(a_1 + 3d) = 7a_1 + 21dWe know that S_7 = 7, so:7a_1 + 21d = 7Let me write that as Equation 2:Equation 2: 7a_1 + 21d = 7Okay, so now we have two equations:1) 2a_1 + 14d = 162) 7a_1 + 21d = 7Hmm, so we have a system of two equations with two variables, a_1 and d. Let's solve this system.First, maybe we can simplify both equations to make it easier.Looking at Equation 1: 2a_1 + 14d = 16We can divide both sides by 2 to simplify:a_1 + 7d = 8Let me call this Equation 1a:Equation 1a: a_1 + 7d = 8Now, Equation 2: 7a_1 + 21d = 7We can factor out 7 from the left side:7(a_1 + 3d) = 7Divide both sides by 7:a_1 + 3d = 1Let me call this Equation 2a:Equation 2a: a_1 + 3d = 1So now, our system is simplified to:1a) a_1 + 7d = 82a) a_1 + 3d = 1Now, let's subtract Equation 2a from Equation 1a to eliminate a_1.So, (a_1 + 7d) - (a_1 + 3d) = 8 - 1Simplify:a_1 + 7d - a_1 - 3d = 7Which simplifies to:4d = 7So, d = 7/4Hmm, so the common difference d is 7/4. Let me note that down.Now, let's plug this value of d back into one of the equations to find a_1. Let's use Equation 2a: a_1 + 3d = 1Substitute d = 7/4:a_1 + 3*(7/4) = 1Compute 3*(7/4): that's 21/4So,a_1 + 21/4 = 1Subtract 21/4 from both sides:a_1 = 1 - 21/4Convert 1 to 4/4:a_1 = 4/4 - 21/4 = (4 - 21)/4 = (-17)/4So, a_1 is -17/4.Alright, so now we have a_1 = -17/4 and d = 7/4.Now, the question is asking for a_{12}. Let's write the formula for a_{12}:a_{12} = a_1 + (12 - 1)d = a_1 + 11dWe have a_1 and d, so plug them in:a_{12} = (-17/4) + 11*(7/4)Compute 11*(7/4): 77/4So,a_{12} = (-17/4) + (77/4) = ( -17 + 77 ) / 4 = 60/4 = 15So, a_{12} is 15.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, a_1 = -17/4, d = 7/4.Compute a_7: a_1 + 6d = (-17/4) + 6*(7/4) = (-17/4) + (42/4) = (25/4)Similarly, a_9 = a_1 + 8d = (-17/4) + 8*(7/4) = (-17/4) + (56/4) = (39/4)So, a_7 + a_9 = 25/4 + 39/4 = 64/4 = 16. That's correct, matches the given information.Now, S_7: sum of first 7 terms.Using the formula S_7 = 7/2*(a_1 + a_7) = 7/2*( (-17/4) + (25/4) ) = 7/2*(8/4) = 7/2*(2) = 7. That's correct too.So, the calculations seem consistent.Therefore, a_{12} is indeed 15.**Final Answer**The value of ( a_{12} ) is boxed{15}.

answer：Alright, so I have this problem: Find all positive integers ( m ) and ( n ) such that ( 2^m - 3^n = 7 ). Hmm, okay. Let me try to figure this out step by step.First, I need to find positive integers ( m ) and ( n ) where the difference between ( 2^m ) and ( 3^n ) is exactly 7. That means ( 2^m ) has to be a bit larger than ( 3^n ). Maybe I can start by plugging in some small values for ( n ) and see if ( 2^m ) becomes a power of 2 that's 7 more than ( 3^n ).Let me write down the equation again:( 2^m = 3^n + 7 )So, I need to find ( m ) and ( n ) such that when I add 7 to ( 3^n ), I get a power of 2. Let's try small values for ( n ) first.Starting with ( n = 1 ):( 3^1 = 3 )So, ( 2^m = 3 + 7 = 10 )But 10 isn't a power of 2. The powers of 2 are 2, 4, 8, 16, 32, etc. So, no solution here.Next, ( n = 2 ):( 3^2 = 9 )So, ( 2^m = 9 + 7 = 16 )Ah, 16 is a power of 2! Specifically, ( 2^4 = 16 ). So, ( m = 4 ) and ( n = 2 ) is a solution.Let me check that: ( 2^4 - 3^2 = 16 - 9 = 7 ). Yep, that works.Okay, so that's one solution. Let's see if there are more.Trying ( n = 3 ):( 3^3 = 27 )So, ( 2^m = 27 + 7 = 34 )34 isn't a power of 2. The next power after 32 is 64, which is way higher. So, no solution here.Moving on to ( n = 4 ):( 3^4 = 81 )So, ( 2^m = 81 + 7 = 88 )88 isn't a power of 2 either. The closest powers are 64 and 128. 88 is in between, so no solution.Next, ( n = 5 ):( 3^5 = 243 )So, ( 2^m = 243 + 7 = 250 )250 isn't a power of 2. Powers of 2 go 128, 256, so 250 is between them. Not a power.Hmm, ( n = 6 ):( 3^6 = 729 )So, ( 2^m = 729 + 7 = 736 )736 isn't a power of 2. The next power after 512 is 1024, so 736 is in between. Not a power.Wait, maybe I should check for ( n = 0 ) just in case, but the problem says positive integers, so ( n ) has to be at least 1. So, ( n = 0 ) is out of the question.Alternatively, maybe I can approach this from the other side. Let's fix ( m ) and see if ( 3^n = 2^m - 7 ) is a power of 3.Starting with small ( m ):( m = 1 ):( 2^1 = 2 )( 3^n = 2 - 7 = -5 ). Negative, so no solution.( m = 2 ):( 2^2 = 4 )( 3^n = 4 - 7 = -3 ). Still negative. No solution.( m = 3 ):( 2^3 = 8 )( 3^n = 8 - 7 = 1 )1 is ( 3^0 ), but ( n ) has to be positive, so no solution here.( m = 4 ):( 2^4 = 16 )( 3^n = 16 - 7 = 9 )9 is ( 3^2 ), so ( n = 2 ). That's the solution we found earlier.( m = 5 ):( 2^5 = 32 )( 3^n = 32 - 7 = 25 )25 isn't a power of 3. Powers of 3 are 3, 9, 27, 81, etc. 25 is between 9 and 27, so no solution.( m = 6 ):( 2^6 = 64 )( 3^n = 64 - 7 = 57 )57 isn't a power of 3. Next power after 27 is 81, so 57 is in between. No solution.( m = 7 ):( 2^7 = 128 )( 3^n = 128 - 7 = 121 )121 is 11 squared, not a power of 3. So, no.( m = 8 ):( 2^8 = 256 )( 3^n = 256 - 7 = 249 )249 isn't a power of 3. The next power after 243 is 729, so 249 is in between. No solution.( m = 9 ):( 2^9 = 512 )( 3^n = 512 - 7 = 505 )505 isn't a power of 3. Powers of 3 near 505 are 243 and 729. So, no.( m = 10 ):( 2^{10} = 1024 )( 3^n = 1024 - 7 = 1017 )1017 isn't a power of 3. The next power after 729 is 2187, so 1017 is in between. No solution.Hmm, seems like as ( m ) increases, ( 2^m ) grows much faster than ( 3^n ). So, the difference ( 2^m - 3^n ) will get larger. But since we're looking for a difference of exactly 7, maybe there's only a limited number of solutions.Wait, but just to be thorough, let me check ( n = 7 ):( 3^7 = 2187 )So, ( 2^m = 2187 + 7 = 2194 )2194 isn't a power of 2. The closest powers are 2048 and 4096. So, no.Similarly, ( n = 8 ):( 3^8 = 6561 )( 2^m = 6561 + 7 = 6568 )6568 isn't a power of 2. The closest is 8192. So, no.I think it's safe to say that as ( n ) increases, ( 3^n ) becomes too large, and ( 2^m ) would have to be enormous to just be 7 more than ( 3^n ). So, likely, the only solution is ( m = 4 ) and ( n = 2 ).But just to make sure I haven't missed anything, let me think about modulo operations. Maybe I can find some constraints on ( m ) and ( n ) using modular arithmetic.Looking at the equation ( 2^m - 3^n = 7 ), let's consider it modulo 3.So, modulo 3:( 2^m equiv 7 mod 3 )But 7 mod 3 is 1, so:( 2^m equiv 1 mod 3 )Now, 2 mod 3 is 2, so 2^1 mod 3 = 2, 2^2 mod 3 = 4 mod 3 = 1, 2^3 mod 3 = 8 mod 3 = 2, 2^4 mod 3 = 16 mod 3 = 1, etc. So, the pattern cycles every 2: 2, 1, 2, 1,...So, ( 2^m equiv 1 mod 3 ) when ( m ) is even. Therefore, ( m ) must be even.Similarly, let's consider the equation modulo 4.( 2^m - 3^n equiv 7 mod 4 )7 mod 4 is 3, so:( 2^m - 3^n equiv 3 mod 4 )Now, 2^m mod 4: For ( m geq 2 ), 2^m is 0 mod 4. For ( m = 1 ), it's 2 mod 4.Similarly, 3^n mod 4: 3 mod 4 is 3, so 3^1 mod 4 = 3, 3^2 mod 4 = 1, 3^3 mod 4 = 3, 3^4 mod 4 = 1, etc. So, it cycles every 2: 3, 1, 3, 1,...So, let's analyze:Case 1: ( m = 1 )Then, ( 2^1 = 2 ). So, equation becomes ( 2 - 3^n = 7 ) which implies ( 3^n = -5 ). Not possible, since ( 3^n ) is positive.Case 2: ( m geq 2 )So, ( 2^m equiv 0 mod 4 ). Therefore, the equation becomes:( 0 - 3^n equiv 3 mod 4 )Which simplifies to:( -3^n equiv 3 mod 4 )Multiply both sides by -1:( 3^n equiv -3 mod 4 )But -3 mod 4 is 1, so:( 3^n equiv 1 mod 4 )From earlier, 3^n mod 4 cycles between 3 and 1. So, ( 3^n equiv 1 mod 4 ) when ( n ) is even.Therefore, ( n ) must be even.So, summarizing the modular results:- ( m ) must be even.- ( n ) must be even.So, both ( m ) and ( n ) are even numbers.Let me denote ( m = 2k ) and ( n = 2l ), where ( k ) and ( l ) are positive integers.So, substituting back into the original equation:( 2^{2k} - 3^{2l} = 7 )Which can be written as:( (2^k)^2 - (3^l)^2 = 7 )This is a difference of squares, so it factors as:( (2^k - 3^l)(2^k + 3^l) = 7 )Now, 7 is a prime number, so its only positive integer factors are 1 and 7.Therefore, we have two possibilities:1. ( 2^k - 3^l = 1 ) and ( 2^k + 3^l = 7 )2. ( 2^k - 3^l = 7 ) and ( 2^k + 3^l = 1 )But the second case is impossible because ( 2^k + 3^l ) is clearly greater than ( 2^k - 3^l ), and both are positive integers. So, the second case would imply ( 2^k + 3^l = 1 ), which is impossible since ( 2^k ) and ( 3^l ) are at least 2 and 3 respectively, so their sum is at least 5.Therefore, only the first case is possible:( 2^k - 3^l = 1 )( 2^k + 3^l = 7 )Let me add these two equations:( (2^k - 3^l) + (2^k + 3^l) = 1 + 7 )Simplifies to:( 2 times 2^k = 8 )So,( 2^{k + 1} = 8 )Which implies:( 2^{k + 1} = 2^3 )Therefore,( k + 1 = 3 )So,( k = 2 )Now, substitute back into one of the equations to find ( l ). Let's take ( 2^k - 3^l = 1 ):( 2^2 - 3^l = 1 )Which is:( 4 - 3^l = 1 )Therefore,( 3^l = 3 )So,( l = 1 )Therefore, ( k = 2 ) and ( l = 1 ), which means:( m = 2k = 4 )( n = 2l = 2 )So, that's the solution we found earlier: ( m = 4 ), ( n = 2 ).But wait, let me just check if there are any other factorizations. Since 7 is prime, the only factor pairs are (1,7) and (-1,-7). But since ( 2^k ) and ( 3^l ) are positive, both ( 2^k - 3^l ) and ( 2^k + 3^l ) must be positive. So, the only possible factorization is (1,7). Therefore, there are no other solutions.Just to be thorough, let me consider if ( k ) or ( l ) could be zero, but since ( m ) and ( n ) are positive integers, ( k ) and ( l ) must be at least 1. So, no issues there.Another thought: could there be a case where ( 2^k - 3^l = -1 ) and ( 2^k + 3^l = -7 )? But since both ( 2^k ) and ( 3^l ) are positive, their sum can't be negative. So, that's impossible.Therefore, the only solution is ( m = 4 ) and ( n = 2 ).Just to recap:- I started by testing small values of ( n ) and ( m ), found a solution at ( m = 4 ), ( n = 2 ).- Then, I used modular arithmetic to constrain ( m ) and ( n ) to be even.- Factored the equation as a difference of squares, leading to a system of equations.- Solved the system to confirm the only solution is ( m = 4 ), ( n = 2 ).I think that's solid. I don't see any other possibilities, so I'm confident that's the only solution.**Final Answer**The only solution is ( m = 4 ) and ( n = 2 ), so the answer is boxed{m=4} and boxed{n=2}.Wait, the instructions say to put the final answer within boxed{}, but it's two variables. Maybe I should write both in a single box? Or perhaps as ordered pairs.Looking back at the problem statement: "Find all positive integers ( m ) and ( n ) such that ( 2^m - 3^n = 7 )." So, the solution is ( m = 4 ) and ( n = 2 ). Since the question asks for both, maybe I should present them together.Perhaps like this: boxed{(4, 2)}.Yeah, that seems appropriate.**Final Answer**The solution is boxed{(4, 2)}.

answer：First, I'll rewrite the equation to combine all terms on one side:[frac{13t^2 - 34t + 12}{3t - 2} + 5t - 6t + 1 = 0]Simplifying the equation gives:[frac{13t^2 - 34t + 12}{3t - 2} - t + 1 = 0]Next, I'll combine the terms by finding a common denominator:[frac{13t^2 - 34t + 12 - t(3t - 2) + (3t - 2)}{3t - 2} = 0]Expanding and simplifying the numerator:[13t^2 - 34t + 12 - 3t^2 + 2t + 3t - 2 = 10t^2 - 29t + 10]So the equation becomes:[frac{10t^2 - 29t + 10}{3t - 2} = 0]For the fraction to be zero, the numerator must be zero:[10t^2 - 29t + 10 = 0]Using the quadratic formula:[t = frac{29 pm sqrt{841 - 400}}{20} = frac{29 pm sqrt{441}}{20} = frac{29 pm 21}{20}]This gives two solutions:[t = frac{50}{20} = frac{5}{2} quad text{and} quad t = frac{8}{20} = frac{2}{5}]Since we're looking for the largest value of ( t ), the answer is ( frac{5}{2} ).

answer：Okay, so I have this problem here where I'm given that (x - 2)^2 + |y - 3| = 0. Hmm, I remember that squares and absolute values are always non-negative, right? So each of these terms must be zero because their sum is zero. That means (x - 2)^2 = 0 and |y - 3| = 0. Let me solve for x and y. If (x - 2)^2 = 0, then x - 2 = 0, so x = 2. Similarly, |y - 3| = 0 implies y - 3 = 0, so y = 3. Got it, so x is 2 and y is 3. Now, I need to simplify and evaluate the expression: [(x - 2y)(x + 2y) - (x - y)^2 + y(y + 2x)] divided by (-2y). Let me write that down step by step.First, let me expand each part of the numerator separately. Starting with (x - 2y)(x + 2y). I remember that this is a difference of squares, so it should be x^2 - (2y)^2, which is x^2 - 4y^2. Next, I have -(x - y)^2. Let me expand (x - y)^2 first. That's x^2 - 2xy + y^2. So, putting the negative sign in front, it becomes -x^2 + 2xy - y^2.Then, the last term is y(y + 2x). Let me distribute the y: that's y^2 + 2xy.Now, let me combine all these parts together. So, the numerator becomes:(x^2 - 4y^2) + (-x^2 + 2xy - y^2) + (y^2 + 2xy)Let me simplify this step by step. First, combine x^2 - 4y^2 with -x^2 + 2xy - y^2. x^2 - x^2 cancels out, so that's 0. Then, -4y^2 - y^2 is -5y^2. And then we have +2xy.Now, adding the last part, which is y^2 + 2xy. So, adding that to the previous result:-5y^2 + 2xy + y^2 + 2xy.Combine like terms. -5y^2 + y^2 is -4y^2. 2xy + 2xy is 4xy.So, the numerator simplifies to -4y^2 + 4xy.Now, the entire expression is this numerator divided by (-2y). So, let me write that:(-4y^2 + 4xy) / (-2y)I can factor out a common factor in the numerator. Let's see, both terms have a -4y. Wait, actually, let me factor out a 4y:4y(-y + x) / (-2y)Wait, but 4y is positive, and the denominator is negative. Alternatively, I can factor out a -4y:-4y(y - x) / (-2y)Hmm, maybe that's better. Let me see:-4y(y - x) divided by (-2y). So, the numerator is -4y(y - x) and the denominator is -2y. I can cancel out the y terms since y is not zero (because y = 3, which is not zero). So, y cancels out, leaving:-4(y - x) / (-2)Simplify the constants: -4 divided by -2 is 2. So, 2(y - x).Alternatively, I can write it as 2(y - x). But since I have x and y values, maybe I can plug them in now.Wait, let me double-check my factoring. Maybe I made a mistake there.Starting from the numerator: -4y^2 + 4xy. Let's factor out 4y:4y(-y + x) = 4y(x - y). So, the numerator is 4y(x - y). So, 4y(x - y) divided by (-2y). Now, y is not zero, so we can cancel y:4(x - y) / (-2)Which simplifies to 4/(-2) * (x - y) = -2(x - y) = -2x + 2y.Alternatively, that's 2y - 2x.Wait, so I have two different expressions here. One is 2(y - x) and another is 2y - 2x. They are the same, right? Because 2(y - x) is 2y - 2x. So, that's consistent.But let me make sure I didn't make a mistake in the factoring.Original numerator: -4y^2 + 4xy.Factor out 4y: 4y(-y + x) = 4y(x - y). Correct.Divide by (-2y): 4y(x - y)/(-2y) = (4/-2)*(x - y) = -2(x - y) = 2y - 2x. Correct.Alternatively, if I factor out -4y: -4y(y - x). Then, divided by (-2y): (-4y)/( -2y) = 2, so 2(y - x). Which is the same as 2y - 2x.So, either way, I get 2y - 2x.Now, since I know x = 2 and y = 3, I can plug those values in.So, 2y - 2x = 2*3 - 2*2 = 6 - 4 = 2.Wait, so the entire expression simplifies to 2.Let me verify this by plugging the values into the original expression without simplifying first.Original expression: [(x - 2y)(x + 2y) - (x - y)^2 + y(y + 2x)] / (-2y)Plugging in x = 2 and y = 3:First term: (2 - 2*3)(2 + 2*3) = (2 - 6)(2 + 6) = (-4)(8) = -32.Second term: -(2 - 3)^2 = -( -1)^2 = -1.Third term: 3(3 + 2*2) = 3(3 + 4) = 3*7 = 21.So, numerator is -32 -1 +21 = (-32 -1) +21 = -33 +21 = -12.Denominator is -2*3 = -6.So, -12 / (-6) = 2. Yep, same result. So, that's correct.Therefore, the simplified expression evaluates to 2.**Final Answer**boxed{2}

answer：Okay, so I have this function f(x) = (a x)/(x + 1), and I need to find the constant a such that when I compose f with itself, meaning f(f(x)), it equals x for all x not equal to -1. Hmm, okay. Let me think about how to approach this.First, I know that function composition means plugging one function into another. So, f(f(x)) means I take f(x) and plug it into f again. So, let me write that out step by step.Given f(x) = (a x)/(x + 1), then f(f(x)) would be f evaluated at f(x). So, replacing every x in f(x) with f(x) itself. Let me write that down:f(f(x)) = f((a x)/(x + 1)).Now, to compute this, I substitute (a x)/(x + 1) into f. So, wherever there's an x in the original function, I replace it with (a x)/(x + 1). Therefore, f(f(x)) becomes:f(f(x)) = [a * (a x)/(x + 1)] / [ (a x)/(x + 1) + 1 ].Okay, that looks a bit complicated, but let's simplify it step by step. Let me handle the numerator and the denominator separately.Starting with the numerator: a multiplied by (a x)/(x + 1). So, that would be (a^2 x)/(x + 1).Now, the denominator is (a x)/(x + 1) + 1. To add these two terms, I need a common denominator. The first term is already over (x + 1), and the second term is 1, which can be written as (x + 1)/(x + 1). So, adding them together:Denominator = (a x)/(x + 1) + (x + 1)/(x + 1) = [a x + (x + 1)] / (x + 1).Let me simplify the numerator of the denominator: a x + x + 1. So, that's (a + 1)x + 1.Putting it all together, the denominator of f(f(x)) is [(a + 1)x + 1]/(x + 1).So now, f(f(x)) is [ (a^2 x)/(x + 1) ] divided by [ ( (a + 1)x + 1 ) / (x + 1) ].When you divide two fractions, it's the same as multiplying by the reciprocal. So, this becomes:f(f(x)) = [ (a^2 x)/(x + 1) ] * [ (x + 1)/( (a + 1)x + 1 ) ].I notice that (x + 1) cancels out in the numerator and denominator, so that simplifies nicely:f(f(x)) = (a^2 x) / [ (a + 1)x + 1 ].So, now we have f(f(x)) expressed as (a^2 x) / [ (a + 1)x + 1 ].But the problem states that f(f(x)) should equal x for all x ≠ -1. So, we set up the equation:(a^2 x) / [ (a + 1)x + 1 ] = x.To solve for a, let's cross-multiply to eliminate the denominator. Multiplying both sides by [ (a + 1)x + 1 ] gives:a^2 x = x * [ (a + 1)x + 1 ].Let me expand the right-hand side:a^2 x = x*(a + 1)x + x*1 = (a + 1)x^2 + x.So, now we have:a^2 x = (a + 1)x^2 + x.Let me bring all terms to one side to set the equation to zero:(a + 1)x^2 + x - a^2 x = 0.Simplify the terms:(a + 1)x^2 + (1 - a^2)x = 0.Now, this equation must hold for all x ≠ -1. That means the coefficients of each power of x must be zero. Otherwise, the equation would only hold for specific x values, not all x.So, let's set the coefficients equal to zero:1. Coefficient of x^2: (a + 1) = 0.2. Coefficient of x: (1 - a^2) = 0.Let me solve the first equation:a + 1 = 0 ⇒ a = -1.Now, let's check the second equation with a = -1:1 - (-1)^2 = 1 - 1 = 0.So, both coefficients are zero when a = -1. That means the equation holds for all x ≠ -1, which is exactly what we need.Wait, hold on a second. Let me double-check my steps because sometimes when dealing with function composition, especially rational functions, there might be something I missed.Starting from f(f(x)) = x, which implies that f is its own inverse function. So, f(f(x)) = x suggests that f is an involution, meaning f is its own inverse. So, perhaps another way to approach this is to set f(f(x)) equal to x and solve for a.But I think my earlier steps are correct. Let me recap:1. Composed f with itself, substituted correctly.2. Simplified the expression step by step, canceling (x + 1) terms.3. Set the resulting expression equal to x.4. Cross-multiplied and expanded, leading to a quadratic in x.5. Since the equation must hold for all x ≠ -1, set coefficients to zero.6. Solved for a, found a = -1, and verified it satisfies both coefficient equations.So, it seems consistent. Just to be thorough, let me plug a = -1 back into the original function and compute f(f(x)) to ensure it indeed equals x.Original function with a = -1:f(x) = (-1 * x)/(x + 1) = -x/(x + 1).Compute f(f(x)):First, f(x) = -x/(x + 1).Then, f(f(x)) = f(-x/(x + 1)).Compute f(-x/(x + 1)):f(-x/(x + 1)) = [ -1 * (-x/(x + 1)) ] / [ (-x/(x + 1)) + 1 ].Simplify numerator:-1 * (-x/(x + 1)) = x/(x + 1).Denominator:(-x/(x + 1)) + 1 = (-x + (x + 1))/(x + 1) = (-x + x + 1)/(x + 1) = 1/(x + 1).So, f(f(x)) = [x/(x + 1)] / [1/(x + 1)] = [x/(x + 1)] * [(x + 1)/1] = x.Yes, that works out perfectly. So, f(f(x)) = x when a = -1.Therefore, the constant a is -1.**Final Answer**The constant ( a ) is boxed{-1}.