🧠:[<think>Okay, so I need to figure out the relationship between y and x for a rectangle with area 12. Let's start by recalling the formula for the area of a rectangle. I think it's length multiplied by width. In this case, one side is x and the other is y, so the area should be x times y. The problem says the area is 12, so maybe I can write that as an equation: x * y = 12. Hmm, but the question is asking for the function relationship between y and x. That usually means expressing y in terms of x, right? So if I have x * y = 12, I can solve for y. Let me try that. Dividing both sides of the equation by x should isolate y. So y = 12 / x. Wait, but is that all? Let me check if there's anything else I need to consider. The problem mentions a rectangle, so the lengths x and y must be positive numbers. So x has to be greater than 0, and y will also be greater than 0. But maybe the question just wants the functional form without specifying the domain. Let me re-read the problem."The area of a rectangle is 12. One side is of length x, and the other side is of length y. The function relationship between y and x is ______."Yes, so they just want how y relates to x as a function. Since we derived y = 12 / x, that should be the answer. Let me think if there's another way to represent it. Maybe writing it as y = 12x^{-1}? But probably the simplest form is y = 12 over x. Is there a chance I made a mistake in the area formula? Let me confirm. Area of rectangle is indeed length times width. If one side is x and the adjacent side is y, then area is x*y. So 12 = x*y, hence y = 12/x. Yep, that seems right.Wait, another thought: sometimes problems specify whether it's a function of x or y. Here they say "the function relationship between y and x", which usually implies y as a function of x. So expressing y in terms of x is correct. If it were x as a function of y, it would be x = 12/y, but the question states y in terms of x. Okay, so I think that's solid. The relationship is y = 12 divided by x. Let me write that again: y = 12/x. That's a hyperbola, but the problem doesn't need the graph, just the function. I don't think there are any other factors here. The problem is straightforward—area given, sides are variables x and y. So the functional relationship is inverse variation. Yep, inverse variation where y varies inversely with x, with the constant of proportionality being 12. So final answer should be y = 12/x. Let me check once more. If x is 3, then y is 4, area is 12. If x is 2, y is 6. That works. If x is 12, y is 1. All positive values. Makes sense. No mistakes here. I think that's correct.Final AnswerThe function relationship between ( y ) and ( x ) is boxed{y = dfrac{12}{x}}.</think>The area of a rectangle is given by the product of its length and width. For a rectangle with sides of length ( x ) and ( y ), the area is ( x times y ). Given that the area is 12, we can write the equation:[ x times y = 12 ]To find the function relationship between ( y ) and ( x ), we solve for ( y ) in terms of ( x ). Dividing both sides of the equation by ( x ), we get:[ y = frac{12}{x} ]Thus, the function relationship between ( y ) and ( x ) is (boxed{y = dfrac{12}{x}}).,<think>Okay, so I have this problem here: The area of a rectangle is 12. One side is of length x, and the other side is of length y. I need to figure out the function relationship between y and x. Hmm, let me think. First, I remember that the area of a rectangle is calculated by multiplying its length and width. So, the formula for area is Area = length × width. In this case, the sides are x and y, so I guess that means Area = x × y. The problem says the area is 12, so substituting that in, I get 12 = x × y. Wait, the question is asking for the function relationship between y and x. So, that means I need to express y as a function of x, right? So starting from the equation 12 = x × y, I need to solve for y. To do that, I can divide both sides of the equation by x. Let me write that out: Starting with 12 = x × y Divide both sides by x: 12/x = ySo, y = 12/x. That seems straightforward. Let me check if that makes sense. If x is, say, 3, then y would be 12/3 = 4. Then the area would be 3 × 4 = 12, which matches. If x is 2, then y is 6, and 2 × 6 is 12. Okay, that works. But wait, do I need to consider any restrictions on x? Like, x can't be zero because you can't have a rectangle with a side length of zero. Also, x has to be positive since lengths can't be negative. So, the domain of x would be x > 0. Similarly, y would also have to be positive. But the problem just asks for the function relationship, so maybe stating y = 12/x is sufficient. Let me make sure there's no other way to interpret the question. The problem says "the function relationship between y and x", so in algebra terms, that means express y in terms of x, which we did. Another way to write the relationship is f(x) = 12/x, where y is a function of x. So either form should be acceptable. Since the question doesn't specify the notation, just giving the equation y = 12/x is probably okay. Is there any chance I made a mistake in rearranging the equation? Let's see: original area is 12 = x*y. Solving for y, divide both sides by x. 12 divided by x is y. Yeah, that's correct. Let me think of another example. If x is 4, then y is 12/4 = 3. Area is 4*3=12. Correct. If x is 1, then y is 12. 1*12=12. Correct. If x is 12, y is 1. 12*1=12. Yep. All these check out. What if x is a fraction? Let's say x is 0.5. Then y is 12/0.5 = 24. Area is 0.5*24=12. Still correct. So even if x is a fractional value, it works. Therefore, regardless of the value of x (as long as it's positive), y is 12 divided by x. So the relationship between y and x is y equals 12 over x. I don't think there's anything else to consider here. The problem is pretty straightforward. The key was remembering the formula for the area of a rectangle and then solving for the desired variable. So, final answer should be y = 12/x.Final AnswerThe function relationship between ( y ) and ( x ) is boxed{y = dfrac{12}{x}}.</think>The area of a rectangle is given by the product of its length and width. Here, the sides are ( x ) and ( y ), and the area is 12. Therefore, we start with the equation:[12 = x times y]To express ( y ) as a function of ( x ), we solve for ( y ) by dividing both sides of the equation by ( x ):[y = frac{12}{x}]We verify this relationship with examples and confirm that it holds true for positive values of ( x ) and ( y ). Thus, the function relationship between ( y ) and ( x ) is:[boxed{y = dfrac{12}{x}}]]

🔑:Given that the area of a rectangle is 12, with one side of length x and the other side of length y, we can express the area as the product of its sides. Therefore, we have:1. The area of the rectangle is given by the product of its sides, which can be written as: [ text{Area} = x times y ]2. Since the area is 12, we substitute this value into the equation: [ 12 = x times y ]3. To find the function relationship between y and x, we solve for y: [ y = frac{12}{x} ]Thus, the function relationship between y and x is boxed{y=frac{12}{x}}.

🧠:[<think>Okay, let's try to tackle this arithmetic sequence problem. Hmm, arithmetic sequences have a common difference, right? So, each term is the previous term plus some constant d. The general formula for the nth term of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.The problem gives us a few pieces of information. First, a_4 + a_7 + a_10 = 17. Second, the sum from a_4 to a_14 inclusive is 77. Third, the last term a_k = 13. We need to find k from the options given.Let me break this down step by step.Starting with the first equation: a_4 + a_7 + a_10 = 17. Let's express each of these terms using the general formula.a_4 is the 4th term: a_1 + 3da_7 is the 7th term: a_1 + 6da_10 is the 10th term: a_1 + 9dSo adding these together:(a_1 + 3d) + (a_1 + 6d) + (a_1 + 9d) = 17Combine like terms:3a_1 + (3d + 6d + 9d) = 173a_1 + 18d = 17Let me note that as equation (1): 3a_1 + 18d = 17Now, the second part says the sum from a_4 to a_14 inclusive is 77. Let's figure out how many terms that is. From term 4 to term 14, inclusive, that's 14 - 4 + 1 = 11 terms.The sum of an arithmetic sequence can be calculated by the formula S = n/2 * (first term + last term). So here, the first term is a_4 and the last term is a_14, and there are 11 terms.So sum = 11/2 * (a_4 + a_14) = 77Let's compute a_4 and a_14:a_4 = a_1 + 3da_14 = a_1 + 13dAdding them together:(a_1 + 3d) + (a_1 + 13d) = 2a_1 + 16dSo the sum equation becomes:11/2 * (2a_1 + 16d) = 77Simplify the left side:11*(a_1 + 8d) = 77Divide both sides by 11:a_1 + 8d = 7Let's call this equation (2): a_1 + 8d = 7Now we have two equations:(1) 3a_1 + 18d = 17(2) a_1 + 8d = 7Let me solve these two equations simultaneously. Maybe we can use substitution or elimination. Let's try elimination.Multiply equation (2) by 3 to match the coefficients of a_1:3*(a_1 + 8d) = 3*7 => 3a_1 + 24d = 21Now subtract equation (1) from this new equation:(3a_1 + 24d) - (3a_1 + 18d) = 21 - 17Simplify:6d = 4Divide both sides by 2:3d = 2 => d = 2/3Wait, 6d = 4, so d = 4/6 = 2/3. Yes, that's correct.Now substitute d = 2/3 into equation (2):a_1 + 8*(2/3) = 7Calculate 8*(2/3) = 16/3So:a_1 + 16/3 = 7Subtract 16/3 from both sides:a_1 = 7 - 16/3 = 21/3 - 16/3 = 5/3So a_1 = 5/3 and d = 2/3.Now we need to find k such that a_k = 13. The formula for a_k is:a_k = a_1 + (k - 1)dSubstitute the known values:13 = 5/3 + (k - 1)*(2/3)Let's solve for k.First, subtract 5/3 from both sides:13 - 5/3 = (k - 1)*(2/3)Convert 13 to thirds: 13 = 39/3So:39/3 - 5/3 = 34/3 = (k - 1)*(2/3)Multiply both sides by 3 to eliminate denominators:34 = (k - 1)*2Divide both sides by 2:17 = k - 1Add 1 to both sides:k = 18Hmm, 18 is one of the answer choices (option B). Let me verify this.Let me check if a_18 is indeed 13.a_1 = 5/3, d = 2/3a_18 = 5/3 + (18 - 1)*(2/3) = 5/3 + 17*(2/3) = 5/3 + 34/3 = 39/3 = 13. Yes, that's correct.Let me also verify the other conditions to make sure.First, check a_4 + a_7 + a_10.a_4 = 5/3 + 3*(2/3) = 5/3 + 6/3 = 11/3 ≈ 3.666...a_7 = 5/3 + 6*(2/3) = 5/3 + 12/3 = 17/3 ≈ 5.666...a_10 = 5/3 + 9*(2/3) = 5/3 + 18/3 = 23/3 ≈ 7.666...Adding them: 11/3 + 17/3 + 23/3 = (11 + 17 + 23)/3 = 51/3 = 17. Correct.Now check the sum from a_4 to a_14 inclusive. First, how many terms? From 4 to 14 inclusive: 14 - 4 + 1 = 11 terms.Sum = (number of terms)/2 * (first term + last term)First term a_4 = 11/3, last term a_14 = 5/3 + 13*(2/3) = 5/3 + 26/3 = 31/3Sum = 11/2 * (11/3 + 31/3) = 11/2 * (42/3) = 11/2 * 14 = (11*14)/2 = 154/2 = 77. Correct.And a_18 = 13 as shown earlier. So all conditions check out.Therefore, the answer is 18, which is option B.Final Answerboxed{B}</think>Given an arithmetic sequence (a_1, a_2, ldots, a_k) with the conditions (a_4 + a_7 + a_{10} = 17), (a_4 + a_5 + cdots + a_{14} = 77), and (a_k = 13), we need to find the value of (k).First, we express the terms (a_4), (a_7), and (a_{10}) using the general formula for an arithmetic sequence (a_n = a_1 + (n-1)d):[begin{aligned}a_4 &= a_1 + 3d, a_7 &= a_1 + 6d, a_{10} &= a_1 + 9d.end{aligned}]Adding these terms together:[(a_1 + 3d) + (a_1 + 6d) + (a_1 + 9d) = 17 implies 3a_1 + 18d = 17. quad text{(1)}]Next, we consider the sum from (a_4) to (a_{14}) inclusive, which is 11 terms. Using the sum formula for an arithmetic sequence:[frac{11}{2} times (a_4 + a_{14}) = 77.]Calculating (a_4) and (a_{14}):[begin{aligned}a_4 &= a_1 + 3d, a_{14} &= a_1 + 13d.end{aligned}]Adding these terms:[(a_1 + 3d) + (a_1 + 13d) = 2a_1 + 16d.]Substituting into the sum formula:[frac{11}{2} times (2a_1 + 16d) = 77 implies 11 times (a_1 + 8d) = 77 implies a_1 + 8d = 7. quad text{(2)}]Solving equations (1) and (2) simultaneously:[begin{aligned}3a_1 + 18d &= 17, a_1 + 8d &= 7.end{aligned}]Multiplying equation (2) by 3:[3a_1 + 24d = 21.]Subtracting equation (1) from this result:[6d = 4 implies d = frac{2}{3}.]Substituting (d = frac{2}{3}) back into equation (2):[a_1 + 8 times frac{2}{3} = 7 implies a_1 + frac{16}{3} = 7 implies a_1 = 7 - frac{16}{3} = frac{5}{3}.]To find (k) such that (a_k = 13):[a_k = a_1 + (k - 1)d implies 13 = frac{5}{3} + (k - 1) times frac{2}{3}.]Solving for (k):[13 - frac{5}{3} = frac{2}{3}(k - 1) implies frac{34}{3} = frac{2}{3}(k - 1) implies 34 = 2(k - 1) implies 17 = k - 1 implies k = 18.]Thus, the value of (k) is (boxed{B}).,<think>Okay, let's see. I need to solve this arithmetic sequence problem. Let's start by recalling what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where each term after the first is obtained by adding a constant difference. So, if the first term is a1 and the common difference is d, then the nth term can be written as a_n = a1 + (n-1)d.The problem gives me a few pieces of information:1. a4 + a7 + a10 = 172. The sum from a4 to a14 inclusive is 77.3. The kth term, a_k, is 13.4. The options for k are 16, 18, 20, 22, 24.I need to find which value of k is correct. Let's break down the information step by step.First, let's express a4, a7, a10, and the other terms in terms of a1 and d.Since a_n = a1 + (n-1)d,a4 = a1 + 3da7 = a1 + 6da10 = a1 + 9dSo, the sum a4 + a7 + a10 = (a1 + 3d) + (a1 + 6d) + (a1 + 9d) = 3a1 + 18dAccording to the problem, this sum is 17. So,3a1 + 18d = 17Let me write that as equation (1):3a1 + 18d = 17.Next, the sum from a4 to a14 inclusive. Let's figure out how many terms that is. From term 4 to term 14, that's 14 - 4 + 1 = 11 terms. So, the sum of 11 terms starting at a4 and ending at a14.To find the sum of an arithmetic sequence, the formula is S = n/2 * (2a1 + (n-1)d), but here we need the sum of terms from a4 to a14. Alternatively, since it's also an arithmetic sequence with the same common difference, we can consider a4 as the first term and then sum 11 terms. The formula would then be S = n/2 * [2a4 + (n -1)d]Alternatively, since a4 is the first term in this sum, the sum from a4 to a14 is equal to the sum of the first 14 terms minus the sum of the first 3 terms. But maybe the first approach is better.Let me try both approaches to check.First approach: sum of 11 terms starting at a4.a4 is the first term, so a4 = a1 + 3d, as before. The common difference is still d. The number of terms is 11. So the sum S = 11/2 * [2a4 + (11 - 1)d] = 11/2 * [2(a1 + 3d) + 10d] = 11/2 * [2a1 + 6d + 10d] = 11/2 * [2a1 + 16d] = 11(a1 + 8d)According to the problem, this sum is 77. So,11(a1 + 8d) = 77Divide both sides by 11:a1 + 8d = 7Let me write this as equation (2):a1 + 8d = 7.Now, we have two equations:(1) 3a1 + 18d = 17(2) a1 + 8d = 7We can solve this system of equations for a1 and d.From equation (2), express a1 in terms of d:a1 = 7 - 8dThen substitute into equation (1):3(7 - 8d) + 18d = 17Calculate:21 - 24d + 18d = 17Combine like terms:21 - 6d = 17Subtract 21 from both sides:-6d = 17 - 21-6d = -4Divide both sides by -6:d = (-4)/(-6) = 4/6 = 2/3So, d = 2/3. Now, substitute back into equation (2):a1 + 8*(2/3) = 7Calculate 8*(2/3) = 16/3So,a1 + 16/3 = 7Subtract 16/3 from both sides:a1 = 7 - 16/3 = 21/3 - 16/3 = 5/3So, a1 = 5/3 and d = 2/3.Now, we know that the kth term is 13. So,a_k = a1 + (k - 1)d = 13Substituting a1 and d:5/3 + (k - 1)*(2/3) = 13Multiply both sides by 3 to eliminate denominators:5 + 2(k - 1) = 39Simplify:5 + 2k - 2 = 39Combine like terms:2k + 3 = 39Subtract 3 from both sides:2k = 36Divide by 2:k = 18Hmm, so k = 18. Let me check if this is one of the answer choices. Yes, option B is 18.Wait, but let me verify to make sure I didn't make any calculation errors.First, check the equations:From a4 + a7 + a10 = 17, with a1 = 5/3 and d = 2/3.a4 = 5/3 + 3*(2/3) = 5/3 + 6/3 = 11/3 ≈ 3.6667a7 = 5/3 + 6*(2/3) = 5/3 + 12/3 = 17/3 ≈ 5.6667a10 = 5/3 + 9*(2/3) = 5/3 + 18/3 = 23/3 ≈ 7.6667Sum: 11/3 + 17/3 + 23/3 = (11 + 17 + 23)/3 = 51/3 = 17. Correct.Then the sum from a4 to a14:Number of terms: 14 - 4 + 1 = 11.Sum S = 11/2 [2*(11/3) + 10*(2/3)] = 11/2 [22/3 + 20/3] = 11/2 * 42/3 = 11/2 * 14 = 11*7 = 77. Correct.Now, check a_k = 13.a1 = 5/3, d = 2/3.a_k = 5/3 + (k -1)*(2/3) = 13.Multiply through by 3:5 + 2(k -1) = 395 + 2k - 2 = 393 + 2k = 392k = 36k = 18. Correct.So, the answer should be 18, which is option B. Let me just confirm if the answer choices are labeled correctly. The options are (A)16, (B)18, (C)20, (D)22, (E)24. Yes, B is 18.Therefore, the answer is B) 18.Final Answerboxed{B}</think>Given an arithmetic sequence (a_1, a_2, ldots, a_k) with the conditions:1. (a_4 + a_7 + a_{10} = 17)2. The sum from (a_4) to (a_{14}) is 77.3. (a_k = 13)We start by expressing the terms in terms of the first term (a_1) and the common difference (d):- (a_4 = a_1 + 3d)- (a_7 = a_1 + 6d)- (a_{10} = a_1 + 9d)From the first condition:[(a_1 + 3d) + (a_1 + 6d) + (a_1 + 9d) = 17 implies 3a_1 + 18d = 17 quad text{(1)}]For the sum from (a_4) to (a_{14}) (11 terms), we use the sum formula for an arithmetic sequence:[text{Sum} = frac{11}{2} left[ 2(a_1 + 3d) + 10d right] = 77 implies 11(a_1 + 8d) = 77 implies a_1 + 8d = 7 quad text{(2)}]Solving the system of equations (1) and (2):- From (2): (a_1 = 7 - 8d)- Substitute into (1): (3(7 - 8d) + 18d = 17 implies 21 - 24d + 18d = 17 implies -6d = -4 implies d = frac{2}{3})- Substitute (d = frac{2}{3}) back into (2): (a_1 + 8 cdot frac{2}{3} = 7 implies a_1 = 7 - frac{16}{3} = frac{5}{3})To find (k) such that (a_k = 13):[a_k = a_1 + (k - 1)d = 13 implies frac{5}{3} + (k - 1) cdot frac{2}{3} = 13]Multiplying through by 3:[5 + 2(k - 1) = 39 implies 2k - 2 = 34 implies 2k = 36 implies k = 18]Thus, the value of (k) is (boxed{B}).]

🔑:Note that a_7-3d=a_4 and a_7+3d=a_{10} where d is the common difference, so a_4+a_7+a_{10}=3a_7=17, or a_7=frac{17}{3}.Likewise, we can write every term in the second equation in terms of a_9, giving us 11a_9=77implies a_9=7.Then the common difference is frac{2}{3}. Then a_k-a_9=13-7=6=9cdotfrac{2}{3}.This means a_k is 9 terms after a_9, so k=18impliesboxed{B}

🧠:[<think>Okay, so I need to find the remainder when the polynomial ( x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) is divided by ( (x - 1) ). Hmm, I remember something about remainders and polynomial division. Let me think. Wait, there's the Remainder Theorem, right? The theorem says that the remainder of a polynomial ( f(x) ) divided by ( (x - a) ) is just ( f(a) ). So in this case, since we're dividing by ( (x - 1) ), the remainder should be ( f(1) ). That seems straightforward. Let me verify that. If I substitute ( x = 1 ) into the polynomial, I get ( 1 + 1^3 + 1^9 + 1^{27} + 1^{81} + 1^{243} ). Since any number to any power is still 1, each term is 1. How many terms are there? Let's count: x, x^3, x^9, x^27, x^81, x^243. That's 6 terms. So 6 times 1 is 6. Therefore, the remainder should be 6. But wait, let me make sure I didn't skip any steps or make a mistake here. Let me recall the Remainder Theorem again. Yes, it states that the remainder is the value of the polynomial evaluated at ( x = a ), which here is 1. So substituting 1 into the polynomial gives the remainder. But just to double-check, maybe I can try doing the polynomial division manually. Let's see. If I divide ( f(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) by ( (x - 1) ), the remainder should be a constant because the divisor is linear. Let me write ( f(x) = (x - 1)Q(x) + R ), where ( Q(x) ) is the quotient and ( R ) is the remainder. Then, substituting ( x = 1 ) into both sides gives ( f(1) = 0 cdot Q(1) + R ), so ( R = f(1) ), which matches the Remainder Theorem. Alternatively, I can perform the division step by step, but that would be tedious for such a high-degree polynomial. Each term in ( f(x) ) is ( x^{3^k} ) where ( k ) ranges from 0 to 5. So, the exponents are 1, 3, 9, 27, 81, 243. Each of these terms can be divided by ( (x - 1) ). But maybe let's consider one term at a time. Let's take the first term, which is ( x ). Divided by ( (x - 1) ), the remainder is ( x ) evaluated at ( x = 1 ), which is 1. Similarly, the second term ( x^3 ) would also have a remainder of 1 when divided by ( (x - 1) ), right? So each term, when divided by ( (x - 1) ), leaves a remainder of 1. Then, adding all those remainders together would give 6, which is the total remainder. Wait, but is that accurate? Because when you add the polynomials, the remainder of the sum is the sum of the remainders. Let me confirm that. Suppose I have two polynomials, ( f(x) ) and ( g(x) ), and I divide each by ( (x - 1) ), getting remainders ( R_f ) and ( R_g ). Then, when I add ( f(x) + g(x) ) and divide by ( (x - 1) ), the remainder should be ( R_f + R_g ). Yes, that makes sense because:( f(x) = (x - 1)Q_f(x) + R_f )( g(x) = (x - 1)Q_g(x) + R_g )Adding them:( f(x) + g(x) = (x - 1)(Q_f(x) + Q_g(x)) + (R_f + R_g) )So the remainder is indeed ( R_f + R_g ). Therefore, since each term ( x^{3^k} ) when divided by ( (x - 1) ) leaves a remainder of 1, adding all six terms gives a total remainder of 6. Alternatively, another way to think about this is that each term ( x^{3^k} ) can be written as ( (x - 1)Q_k(x) + 1 ), because when you substitute ( x = 1 ), it's 1. Therefore, adding them all together:( f(x) = sum_{k=0}^{5} x^{3^k} = sum_{k=0}^{5} [(x - 1)Q_k(x) + 1] = (x - 1)sum_{k=0}^{5} Q_k(x) + 6 )Which shows that the remainder is 6. Is there any other way to verify this? Maybe using modular arithmetic for polynomials. If we consider the polynomial modulo ( (x - 1) ), then ( x equiv 1 mod (x - 1) ). Therefore, each term ( x^{3^k} equiv 1^{3^k} = 1 mod (x - 1) ). Therefore, adding six terms each congruent to 1 modulo ( (x - 1) ) gives ( 6 mod (x - 1) ), so the remainder is 6. Yes, this all seems consistent. I can't think of a reason why this would be incorrect. All the different approaches—the Remainder Theorem, polynomial division properties, and modular arithmetic—lead to the same conclusion. Therefore, the remainder should be 6. Final AnswerThe remainder is boxed{6}.</think>To find the remainder when the polynomial ( f(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) is divided by ( (x - 1) ), we use the Remainder Theorem. The theorem states that the remainder of a polynomial ( f(x) ) divided by ( (x - a) ) is ( f(a) ). Here, ( a = 1 ).Substituting ( x = 1 ) into the polynomial:[f(1) = 1 + 1^3 + 1^9 + 1^{27} + 1^{81} + 1^{243}]Since any number to any power is still 1, each term evaluates to 1. There are 6 terms in total, so:[f(1) = 1 + 1 + 1 + 1 + 1 + 1 = 6]Thus, the remainder when ( f(x) ) is divided by ( (x - 1) ) is (boxed{6}).,<think>Okay, let me try to figure out this problem. So, the question is: What remainder does the polynomial ( x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) give when divided by ( (x - 1) )? Hmm, right. I need to find the remainder when this polynomial is divided by ( x - 1 ). First, I remember that when you divide a polynomial by ( x - a ), the remainder is just the value of the polynomial evaluated at ( x = a ). That's the Remainder Theorem, right? So in this case, since we're dividing by ( x - 1 ), the remainder should be the polynomial evaluated at ( x = 1 ). That seems straightforward. Let me check if I remember that correctly. So, if I take the polynomial ( P(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ), then the remainder when divided by ( x - 1 ) should be ( P(1) ). Let me confirm that. Yeah, the Remainder Theorem states that ( P(a) ) is the remainder when ( P(x) ) is divided by ( x - a ). So, substituting ( x = 1 ) into the polynomial should give the remainder. Alright, so let me compute ( P(1) ). Each term in the polynomial is ( x ) raised to some power. So plugging in 1 for x, each term becomes ( 1^n ), where n is the exponent. Since any number to any power is still 1, right? So, ( 1^1 = 1 ), ( 1^3 = 1 ), ( 1^9 = 1 ), and so on. Therefore, each term will just be 1. Let me count how many terms there are. Starting with ( x ), then ( x^3 ), ( x^9 ), ( x^{27} ), ( x^{81} ), ( x^{243} ). That's 1, 2, 3, 4, 5, 6 terms. So each term is 1, and there are 6 terms. Therefore, the sum is 1 + 1 + 1 + 1 + 1 + 1 = 6. So, according to the Remainder Theorem, the remainder should be 6. Therefore, the answer is 6. Wait, but let me make sure there's no catch here. Sometimes problems might look straightforward but have a twist. Let me think again. Is there any reason why the Remainder Theorem wouldn't apply here? The divisor is ( x - 1 ), which is a linear polynomial, so the theorem should hold. The polynomial given is indeed a polynomial in x with integer exponents, so evaluating at x = 1 is valid. Alternatively, maybe I can do the division manually to check. Let's consider dividing ( x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) by ( x - 1 ). But that seems tedious given the high exponents. But maybe using polynomial long division concepts. Alternatively, express the polynomial as ( Q(x)(x - 1) + R ), where R is the remainder. Since we're dividing by a linear polynomial, the remainder should be a constant. Then, setting x = 1, we get ( P(1) = Q(1)(1 - 1) + R ), which simplifies to ( P(1) = R ). So again, R = P(1) = 6. Therefore, regardless of the approach, the remainder is 6. Alternatively, maybe use synthetic division. Let me try synthetic division with root 1. For synthetic division, we write the coefficients of the polynomial. But the polynomial here is sparse; it only has terms with exponents 1, 3, 9, 27, 81, 243. So, in terms of coefficients, for each exponent from 0 up to 243, most coefficients are 0 except for exponents 1,3,9,27,81,243, where the coefficient is 1. But synthetic division with such a high degree polynomial would be impractical. For example, writing out 244 coefficients (from x^0 to x^243) would take too long, and most of them are zero. So, this method isn't practical here. Alternatively, think of the polynomial as a sum of geometric progressions? Wait, but each exponent is 3 raised to the power of the term's position. Let's see:The exponents are 1, 3, 9, 27, 81, 243. Notice that each exponent is 3 times the previous one: 1, 3 (1*3), 9 (3*3), 27 (9*3), 81 (27*3), 243 (81*3). So the exponents are 3^0, 3^1, 3^2, 3^3, 3^4, 3^5. That's an interesting pattern. So the exponents are 3^k where k ranges from 0 to 5. Therefore, the polynomial can be written as ( sum_{k=0}^{5} x^{3^k} ).But how does that help? Maybe recognizing the pattern isn't necessary here since we're just evaluating at x = 1, which turns each term into 1. So regardless of the exponent, each term is 1 when x = 1, so sum is 6. Alternatively, maybe there's a different approach required, but in this case, the straightforward application of the Remainder Theorem seems sufficient. Wait a second, but let me just confirm that the remainder is indeed a constant. Since we are dividing by a first-degree polynomial, the remainder must be of degree less than 1, which is a constant. Therefore, the remainder is just a number, which we found to be 6. Therefore, unless there's a miscalculation in the number of terms, which there isn't—since 1, 3, 9, 27, 81, 243 are six exponents—the remainder is 6. Alternatively, maybe the problem is trying to trick me into thinking it's 6, but actually, there's some modulus involved? Wait, but when we divide by x - 1, the remainder is over the polynomials, meaning it's an integer (assuming coefficients are integers). Since all coefficients are integers, the remainder will be an integer. Since substituting x = 1 gives 6, the remainder is 6. Therefore, after all these checks, I think the answer is 6. Final Answerboxed{6}</think>To find the remainder when the polynomial ( P(x) = x + x^3 + x^9 + x^{27} + x^{81} + x^{243} ) is divided by ( (x - 1) ), we use the Remainder Theorem. The Remainder Theorem states that the remainder of a polynomial ( P(x) ) divided by ( (x - a) ) is ( P(a) ).Here, we need to evaluate ( P(1) ). Substituting ( x = 1 ) into the polynomial, each term ( x^{3^k} ) (where ( k ) ranges from 0 to 5) becomes ( 1^{3^k} = 1 ). The polynomial has six terms: ( x, x^3, x^9, x^{27}, x^{81}, x^{243} ). Each term evaluates to 1 when ( x = 1 ), so the sum is:[1 + 1 + 1 + 1 + 1 + 1 = 6]Thus, the remainder when ( P(x) ) is divided by ( (x - 1) ) is (boxed{6}).]

🔑:10.31. Answer: 6. Let x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=P(x) times times(x-1)+r. By setting x=1, we get r=6.

🧠:[<think>Okay, so I need to solve this problem about the function f(x) and then apply it to a triangle problem. Let me start with part (1), finding the monotonic intervals of f(x). The function is given as:f(x) = sin(x + 5π/2) cos(x - π/2) - cos²(x + π/4)First, maybe I should simplify this expression using trigonometric identities to make it easier to analyze. Let's see.Starting with sin(x + 5π/2). Remember that sine has a period of 2π, so adding multiples of 2π doesn't change the value. Let's subtract 2π (which is 4π/2) from 5π/2, so 5π/2 - 4π/2 = π/2. So sin(x + 5π/2) is the same as sin(x + π/2). And sin(x + π/2) is equal to cos(x), right? Because sin(θ + π/2) = cosθ. So that part simplifies to cos(x).Next, cos(x - π/2). Similarly, cosine of (θ - π/2) is sinθ because cos(θ - π/2) = sinθ. So cos(x - π/2) is sin(x).So the first term simplifies to cos(x) * sin(x). So that's (1/2) sin(2x) maybe? Because sin(2x) = 2 sinx cosx, so sinx cosx = (1/2) sin2x. But let me check that later.Now the second term is -cos²(x + π/4). Let me see if I can simplify that. Using the identity cos²θ = (1 + cos2θ)/2. So cos²(x + π/4) = [1 + cos(2x + π/2)] / 2. Then multiplied by -1, so - [1 + cos(2x + π/2)] / 2.Putting it all together:f(x) = sin(x + 5π/2) cos(x - π/2) - cos²(x + π/4) = cosx sinx - [1 + cos(2x + π/2)] / 2Simplify:First term: sinx cosx = (1/2) sin2x.Second term: - [1 + cos(2x + π/2)] / 2.Now, cos(2x + π/2) can be simplified. Cos(θ + π/2) = -sinθ. So cos(2x + π/2) = -sin2x. Therefore, the second term becomes:- [1 + (-sin2x)] / 2 = - (1 - sin2x)/2 = -1/2 + (sin2x)/2.So putting it all together:f(x) = (1/2) sin2x - 1/2 + (1/2) sin2x = (1/2 sin2x + 1/2 sin2x) - 1/2 = sin2x - 1/2.Wait, that simplifies nicely! Let me verify that step by step.Original expression:f(x) = (1/2 sin2x) - [1 + cos(2x + π/2)] / 2Then, as we saw, cos(2x + π/2) = -sin2x, so:f(x) = (1/2 sin2x) - [1 - sin2x]/2= (1/2 sin2x) - 1/2 + (1/2 sin2x)Combine like terms:1/2 sin2x + 1/2 sin2x = sin2xThen subtract 1/2:f(x) = sin2x - 1/2Wow, that's much simpler! So the function simplifies to sin2x - 1/2. That makes the problem much easier. So instead of dealing with those phase shifts and squares, it's just sin2x minus a constant. Great.Now, for part (1), find the monotonic intervals of f(x). Since f(x) is sin2x - 1/2, its derivative is 2cos2x. So f'(x) = 2cos2x. To find monotonic intervals, we need to find where f'(x) is positive or negative.So f'(x) = 2cos2x. The critical points occur where cos2x = 0. Solving cos2x = 0:2x = π/2 + kπ, where k is any integer.So x = π/4 + kπ/2.Therefore, the critical points are at x = π/4, 3π/4, 5π/4, etc.Now, we need to analyze the intervals between these critical points to see where the derivative is positive or negative.Let's consider the intervals:1. Between x = -π/4 and x = π/4: Let's pick x = 0. Then cos(0) = 1, so f'(0) = 2*1 = 2 > 0. So increasing.2. Between x = π/4 and x = 3π/4: Let's take x = π/2. Then cos(2*(π/2)) = cosπ = -1. So f'(π/2) = 2*(-1) = -2 < 0. Decreasing.3. Between x = 3π/4 and x = 5π/4: Take x = π. cos2π = 1. So f'(π) = 2*1 = 2 > 0. Increasing.4. Between x = 5π/4 and x = 7π/4: x = 3π/2. cos(3π) = -1. f'(3π/2) = -2 < 0. Decreasing.And so on. So the function alternates between increasing and decreasing every π/2 interval. Therefore, the monotonic intervals are:- Increasing on intervals (-π/4 + kπ, π/4 + kπ) for all integers k.- Decreasing on intervals (π/4 + kπ, 3π/4 + kπ) for all integers k.But the problem just says "find the monotonic intervals of f(x)" without specifying a domain. Assuming the domain is all real numbers, then we can write them as above with k being integers.But maybe the problem expects intervals in terms of a single period. Since the function is periodic with period π (since sin2x has period π). So over each interval of length π, the function increases from -π/4 to π/4, then decreases from π/4 to 3π/4, then increases again? Wait, wait, no. Wait, the critical points are at π/4 + kπ/2. So in each π interval, there are two increasing and two decreasing intervals?Wait, let me check again. The critical points are at x = π/4 + kπ/2. So between each consecutive critical points, the derivative changes sign. So every π/2 interval, the sign changes. But the function's period is π. Hmm.Wait, sin2x has period π. So the entire function f(x) = sin2x - 1/2 has period π. Therefore, the monotonicity repeats every π.Therefore, the increasing intervals would be, for example, from -π/4 to π/4, then decreasing from π/4 to 3π/4, and then increasing again from 3π/4 to 5π/4, but since the period is π, the interval from 3π/4 to 5π/4 is equivalent to -π/4 to π/4 shifted by π. Wait, maybe better to write the intervals within one period, say from -π/4 to 3π/4.Wait, maybe it's clearer to state the intervals in terms of the standard period. Let me choose the interval [0, π] as a representative period.But when x is in [0, π/4], derivative is positive (since 2x is between 0 and π/2, so cos2x positive). Then x in (π/4, 3π/4): 2x is between π/2 and 3π/2, so cos2x is negative, so derivative negative. Then x in (3π/4, π): 2x is between 3π/2 and 2π, so cos2x is positive again, so derivative positive. Wait, but hold on. Wait, when x is in (3π/4, 5π/4), but since our period is π, x can't go beyond π here. Wait, no, 2x when x is in (3π/4, π) is (3π/2, 2π), where cosine is positive (since it's from 3π/2 to 2π, which is the fourth quadrant, so cosine positive). So in that interval, derivative is positive. So in [0, π], the function is increasing on [0, π/4], decreasing on [π/4, 3π/4], and increasing again on [3π/4, π].But since the function is periodic, these intervals repeat every π. So the general monotonic intervals would be:Increasing on intervals ( (-π/4) + kπ, π/4 + kπ )Decreasing on intervals ( π/4 + kπ, 3π/4 + kπ )for all integers k.But usually, when asked for monotonic intervals, they might expect the answer in terms of intervals within a single period, but since no specific interval is given, we need to state all intervals.But perhaps the problem expects intervals without reference to k, but in terms of the principal ones. Let me check.Wait, perhaps the original function f(x) is defined for all real numbers, and we have to state the intervals in terms of where it's increasing or decreasing. So as per the critical points at x = π/4 + kπ/2, the intervals between each pair of consecutive critical points alternate between increasing and decreasing.Therefore, the function is increasing on intervals ( π/4 + (k-1)π/2, π/4 + kπ/2 ) when k is even, and decreasing when k is odd, or something like that. Wait, maybe better to write as:Between each critical point x = π/4 + kπ/2 and x = π/4 + (k+1)π/2, the function is increasing or decreasing depending on k.Wait, perhaps another approach: since the derivative is 2cos2x, which is positive when cos2x is positive, i.e., when 2x is in (-π/2 + 2πn, π/2 + 2πn), so x is in (-π/4 + πn, π/4 + πn) for integer n. Therefore, f(x) is increasing on (-π/4 + πn, π/4 + πn) and decreasing on (π/4 + πn, 3π/4 + πn) for all integers n.Yes, that's more precise. So the increasing intervals are (-π/4 + πn, π/4 + πn), and decreasing intervals are (π/4 + πn, 3π/4 + πn) for all integers n.So part (1) answer is that f(x) is increasing on intervals ( πn - π/4, πn + π/4 ) and decreasing on intervals ( πn + π/4, πn + 3π/4 ) for all integers n.But the problem might want the intervals expressed in a different way. Maybe using union of intervals over the real line. But since it's periodic, it's standard to express it in terms of shifted by π intervals.Okay, so part (1) is solved.Now moving to part (2). In an acute triangle ABC, sides opposite angles A, B, C are a, b, c respectively. Given that f(A/2) = (√3 - 1)/2, and a = 1, find the maximum value of the perimeter of triangle ABC.First, recall that the perimeter is a + b + c = 1 + b + c. So we need to maximize b + c given the constraints.Given f(A/2) = (√3 - 1)/2. Since f(x) = sin2x - 1/2, so substituting x = A/2:f(A/2) = sin(2*(A/2)) - 1/2 = sinA - 1/2 = (√3 - 1)/2.Therefore, sinA - 1/2 = (√3 - 1)/2. Let's solve for sinA.Adding 1/2 to both sides:sinA = (√3 - 1)/2 + 1/2 = (√3 - 1 + 1)/2 = √3/2.Therefore, sinA = √3/2. So angle A is either π/3 or 2π/3. However, since the triangle is acute, all angles are less than π/2. Wait, acute triangle means all angles less than π/2 (90 degrees). Therefore, if sinA = √3/2, then angle A is π/3 (60 degrees), since 2π/3 is 120 degrees, which is obtuse. Therefore, angle A = π/3.So angle A is 60 degrees. Then sides: in triangle ABC, a = 1, opposite angle A = π/3. We need to find the maximum perimeter, i.e., maximize b + c + 1.So to maximize b + c. Since the triangle is acute, all angles are less than π/2. So angles B and C are also less than π/2, and angle A is π/3.We can use the Law of Sines. In triangle ABC:a / sinA = b / sinB = c / sinC = 2R, where R is the circumradius.Given a = 1, angle A = π/3, so:1 / sin(π/3) = 2R => 1 / (√3/2) = 2R => 2/√3 = 2R => R = 1/√3.Therefore, b = 2R sinB = (2/√3) sinB, and c = (2/√3) sinC.So perimeter P = 1 + (2/√3)(sinB + sinC).But in a triangle, angles sum to π. So B + C = π - A = π - π/3 = 2π/3. So C = 2π/3 - B.Therefore, sinC = sin(2π/3 - B) = sin(2π/3)cosB - cos(2π/3)sinB = (√3/2)cosB - (-1/2)sinB = (√3/2)cosB + (1/2)sinB.Therefore, sinB + sinC = sinB + (√3/2)cosB + (1/2)sinB = (3/2)sinB + (√3/2)cosB.Therefore, perimeter P = 1 + (2/√3)[ (3/2)sinB + (√3/2)cosB ].Simplify:(2/√3)*(3/2 sinB) = (2/√3)*(3/2) sinB = (3/√3) sinB = √3 sinB.(2/√3)*(√3/2 cosB) = (2/√3)*(√3/2) cosB = cosB.Therefore, P = 1 + √3 sinB + cosB.So P = 1 + √3 sinB + cosB. Now, we need to maximize this expression with respect to B, under the constraints that the triangle is acute. Since all angles are less than π/2, angle B < π/2, and angle C = 2π/3 - B < π/2.So angle C < π/2 implies 2π/3 - B < π/2 => B > 2π/3 - π/2 = π/6. So angle B must be greater than π/6 and less than π/2. So B ∈ (π/6, π/2). Similarly, since angle C = 2π/3 - B < π/2, which gives B > π/6, and angle B < π/2. So the domain of B is (π/6, π/2).Therefore, we need to maximize P = 1 + √3 sinB + cosB over B ∈ (π/6, π/2).To find the maximum of √3 sinB + cosB. Let's denote this as Q = √3 sinB + cosB. We can write Q as 2 sin(B + π/6), using the identity a sinθ + b cosθ = R sin(θ + φ), where R = √(a² + b²), φ = arctan(b/a).Here, a = √3, b = 1. So R = √( (√3)² + 1² ) = √(3 + 1) = 2. Then φ = arctan(1/√3) = π/6. Therefore:√3 sinB + cosB = 2 sin(B + π/6)Therefore, Q = 2 sin(B + π/6). Therefore, P = 1 + 2 sin(B + π/6).So to maximize P, we need to maximize sin(B + π/6). The maximum value of sin is 1, so if possible, the maximum would be 1 + 2*1 = 3. But we need to check if B + π/6 can reach π/2 where sin is 1, given the domain of B.Given B ∈ (π/6, π/2). So B + π/6 ∈ (π/6 + π/6, π/2 + π/6) = (π/3, 2π/3). The sine function in (π/3, 2π/3) reaches maximum at π/2. So when does B + π/6 = π/2? Then B = π/2 - π/6 = π/3. So B = π/3 is within the domain (π/6, π/2)? π/3 ≈ 1.047, π/6 ≈ 0.523, π/2 ≈ 1.570. Yes, π/3 is within (π/6, π/2). Therefore, the maximum of sin(B + π/6) is 1, achieved when B = π/3.Therefore, maximum P = 1 + 2*1 = 3.But wait, hold on. If B = π/3, then angle C = 2π/3 - π/3 = π/3. So all angles would be π/3, making the triangle equilateral. But in that case, all angles are π/3, which are less than π/2, so the triangle is acute. Therefore, this is allowed.Therefore, the maximum perimeter is 3. But wait, let's verify this.If the triangle is equilateral with all sides equal, then a = 1, so all sides are 1. Then perimeter would be 3, which matches. But in that case, since a = 1, sides opposite 60 degrees, so using the Law of Sines: 1 / sin(60°) = 2R, so 1 / (√3/2) = 2R => 2/√3 = 2R => R = 1/√3. Then sides b and c would be 2R sinB and 2R sinC. If B and C are 60°, then b = c = 2*(1/√3)*sin60° = 2*(1/√3)*(√3/2) = 1. Therefore, perimeter is 1 + 1 + 1 = 3. So that works.But wait, is there a possibility that the maximum occurs at the endpoints of the interval? Let's check the endpoints.When B approaches π/6 from the right:B = π/6 + ε, then angle C = 2π/3 - π/6 - ε = π/2 - ε. So C approaches π/2 from below.Then Q = √3 sinB + cosB. Let's compute when B approaches π/6:sin(π/6) = 1/2, cos(π/6) = √3/2. So Q approaches √3*(1/2) + √3/2 = √3*(1/2 + 1/2) = √3. Therefore, P approaches 1 + √3 ≈ 1 + 1.732 ≈ 2.732.When B approaches π/2 from the left:Angle C approaches 2π/3 - π/2 = π/6. Then Q = √3 sinB + cosB.At B = π/2, sinB = 1, cosB = 0. So Q = √3*1 + 0 = √3 ≈ 1.732. So P approaches 1 + √3 ≈ 2.732 as well.But at B = π/3, Q = 2*1 = 2, so P = 3. So indeed, the maximum is achieved at B = π/3, giving perimeter 3. Therefore, the maximum perimeter is 3.But wait, but in an equilateral triangle, all angles are π/3, which are acute, so the triangle is acute. Therefore, that's valid. So the maximum perimeter is 3.But let me think again. Wait, is there any other constraint? The problem says "acute triangle", so all angles must be less than π/2. If we have an equilateral triangle, all angles are π/3, which is less than π/2, so it is indeed acute. Therefore, the maximum perimeter is 3.But let me confirm with another approach. Suppose we use the Law of Cosines. In triangle ABC, with a = 1, angle A = π/3. Then, using the Law of Cosines:a² = b² + c² - 2bc cosA => 1 = b² + c² - 2bc cos(π/3) => 1 = b² + c² - bc.We need to maximize b + c, given that 1 = b² + c² - bc.Let me set S = b + c, P = bc. Then from the equation:1 = (b + c)² - 3bc = S² - 3P.So S² - 3P = 1 => P = (S² - 1)/3.But also, for real positive numbers b and c, by AM ≥ GM: bc ≤ (b + c)^2 / 4 = S² /4. So P ≤ S² /4.Therefore, (S² - 1)/3 ≤ S² /4.Multiply both sides by 12:4(S² - 1) ≤ 3S² => 4S² - 4 ≤ 3S² => S² - 4 ≤ 0 => S² ≤ 4 => S ≤ 2.But wait, if S ≤ 2, then the maximum possible perimeter is 1 + 2 = 3. Wait, but in our previous approach, we found perimeter 3 when S = 2. Therefore, that's consistent.But here, from the Law of Cosines approach, we see that S (which is b + c) is at most 2. Therefore, the maximum perimeter is 1 + 2 = 3. Therefore, confirming our previous result.Additionally, equality occurs when b = c, since in AM ≥ GM, equality is when b = c. So when b = c, then the triangle is isoceles with b = c, angle B = angle C = π/3, hence equilateral. So this again gives perimeter 3.Therefore, the maximum perimeter is 3.But wait, hold on. The problem says "acute triangle". In the case where angle C approaches π/2, does the expression for perimeter approach 1 + √3? Which is approximately 2.732, which is less than 3. So the maximum occurs at the equilateral triangle, which is indeed acute. Therefore, the answer is 3.But let me check once more with calculus. We had P = 1 + 2 sin(B + π/6). To maximize this, since B ∈ (π/6, π/2), then B + π/6 ∈ (π/3, 2π/3). The maximum of sin in this interval is at π/2, so sin(π/2) = 1. So maximum P = 1 + 2*1 = 3. Therefore, yes, confirmed.Thus, the maximum perimeter is 3.Final Answer(1) The function ( f(x) ) is increasing on intervals ( left( frac{pi}{4} + kpi, frac{3pi}{4} + kpi right) ) for all integers ( k ).Wait, wait, hold on. Wait in the initial analysis, we found that f'(x) = 2cos2x. So when cos2x is positive, f is increasing. Which occurs when 2x ∈ (-π/2 + 2πk, π/2 + 2πk), so x ∈ (-π/4 + πk, π/4 + πk). Therefore, increasing intervals are ( πk - π/4, πk + π/4 ), and decreasing intervals are ( πk + π/4, πk + 3π/4 ) for all integers k.Therefore, my previous conclusion was correct. But in the final answer above, I mistakenly wrote increasing on intervals ( π/4 +kπ, 3π/4 +kπ ), which is actually the decreasing intervals. So that's a mistake.Wait, let me correct. The correct intervals are:Increasing on ( left( -frac{pi}{4} + kpi, frac{pi}{4} + kpi right) )Decreasing on ( left( frac{pi}{4} + kpi, frac{3pi}{4} + kpi right) )for all integers k.Therefore, in the final answer for part (1), it should be:(1) ( f(x) ) is increasing on intervals ( left( kpi - frac{pi}{4}, kpi + frac{pi}{4} right) ) and decreasing on intervals ( left( kpi + frac{pi}{4}, kpi + frac{3pi}{4} right) ) for all integers ( k ).But in the initial thought process, I messed up while writing the final answer. So need to correct that.Similarly, part (2) answer is 3.So final answers:(1) Monotonic intervals: Increasing on ( left( kpi - frac{pi}{4}, kpi + frac{pi}{4} right) ), decreasing on ( left( kpi + frac{pi}{4}, kpi + frac{3pi}{4} right) ) for all integers ( k ).(2) Maximum perimeter is 3.But the problem statement in Chinese might have different interval notation, but according to the calculations, that's correct.Final Answer(1) The monotonic intervals of ( f(x) ) are increasing on ( left( kpi - frac{pi}{4}, kpi + frac{pi}{4} right) ) and decreasing on ( left( kpi + frac{pi}{4}, kpi + frac{3pi}{4} right) ) for all integers ( k ).(2) The maximum value of the perimeter of ( triangle ABC ) is (boxed{3}).</think>Given the function ( f(x) = sin left( x + frac{5pi}{2} right) cos left( x - frac{pi}{2} right) - cos^2 left( x + frac{pi}{4} right) ):1. Monotonic Intervals of ( f(x) ): - Simplify the function using trigonometric identities: [ sin left( x + frac{5pi}{2} right) = cos(x), quad cos left( x - frac{pi}{2} right) = sin(x), quad cos^2 left( x + frac{pi}{4} right) = frac{1 + cos left( 2x + frac{pi}{2} right)}{2} ] [ cos left( 2x + frac{pi}{2} right) = -sin(2x) ] [ f(x) = cos(x) sin(x) - frac{1 - sin(2x)}{2} = sin(2x) - frac{1}{2} ] - The derivative ( f'(x) = 2cos(2x) ). - Critical points where ( cos(2x) = 0 ) occur at ( x = frac{pi}{4} + kfrac{pi}{2} ) for integers ( k ). - Determine intervals where ( f'(x) ) is positive or negative: - Increasing on ( left( kpi - frac{pi}{4}, kpi + frac{pi}{4} right) ) - Decreasing on ( left( kpi + frac{pi}{4}, kpi + frac{3pi}{4} right) ) for all integers ( k ).2. Maximum Perimeter of Acute Triangle ( triangle ABC ): - Given ( fleft( frac{A}{2} right) = frac{sqrt{3} - 1}{2} ): [ sin(A) - frac{1}{2} = frac{sqrt{3} - 1}{2} implies sin(A) = frac{sqrt{3}}{2} implies A = frac{pi}{3} ] - Using the Law of Sines and Law of Cosines: - ( a = 1 ), ( A = frac{pi}{3} ) - Maximize ( b + c ) under the constraint of an acute triangle. - Convert to a single variable optimization problem using angle ( B ): [ P = 1 + sqrt{3} sin(B) + cos(B) ] - Maximum value occurs when ( B = frac{pi}{3} ), leading to an equilateral triangle with perimeter ( 3 ). Final Answer(1) The function ( f(x) ) is increasing on intervals ( left( kpi - frac{pi}{4}, kpi + frac{pi}{4} right) ) and decreasing on intervals ( left( kpi + frac{pi}{4}, kpi + frac{3pi}{4} right) ) for all integers ( k ).(2) The maximum value of the perimeter of ( triangle ABC ) is (boxed{3}).,<think>Okay, so I need to tackle this problem step by step. Let me start with the first part: finding the monotonic intervals of the function f(x) = sin(x + 5π/2)cos(x - π/2) - cos²(x + π/4). Hmm, that looks a bit complicated with all those phase shifts. Maybe I can simplify the expression first using trigonometric identities.Starting with the first term: sin(x + 5π/2). Remember that sine has a period of 2π, so adding 5π/2 is the same as adding π/2 after subtracting 2π. So, sin(x + 5π/2) = sin(x + π/2). But sin(x + π/2) is equal to cos(x), right? Because sin(θ + π/2) = cosθ. So that first part simplifies to cos(x).Next, the second term is cos(x - π/2). Similarly, cosine of (θ - π/2) is sinθ, because cos(θ - π/2) = sinθ. So cos(x - π/2) becomes sin(x). So multiplying the first two terms: cos(x) * sin(x). That simplifies the first part of the function to (1/2) sin(2x) using the double-angle identity sin(2x) = 2 sinx cosx. Wait, but actually, just cosx sinx is (1/2) sin(2x). So the first term simplifies to (1/2) sin(2x).Now the third term is -cos²(x + π/4). Let me recall that cos²θ can be written as (1 + cos2θ)/2. So applying that here: cos²(x + π/4) = [1 + cos(2x + π/2)] / 2. The negative sign in front makes it - [1 + cos(2x + π/2)] / 2. Let me compute cos(2x + π/2). Using the identity cos(θ + π/2) = -sinθ, so cos(2x + π/2) = -sin(2x). Therefore, the third term becomes - [1 - sin(2x)] / 2 = (-1/2) + (sin2x)/2.Putting it all together, the function f(x) is:First term: (1/2) sin2xThird term: (-1/2) + (sin2x)/2So combining these: (1/2) sin2x - 1/2 + (1/2) sin2x = sin2x - 1/2.Wait, that seems too simple. Let me check again.Original function:sin(x + 5π/2) * cos(x - π/2) - cos²(x + π/4)We simplified sin(x +5π/2) to cosx, and cos(x - π/2) to sinx. So first term is cosx sinx = (1/2) sin2x.Second term: -cos²(x + π/4). As above, cos²(x + π/4) = [1 + cos(2x + π/2)] / 2. Then cos(2x + π/2) = -sin2x. Therefore, cos²(x + π/4) = [1 - sin2x]/2. Therefore, -cos²(x + π/4) = - [1 - sin2x]/2 = -1/2 + (sin2x)/2.Therefore, combining both terms:First term: (1/2) sin2xSecond term: -1/2 + (1/2) sin2xSo total f(x) = (1/2 sin2x) + (-1/2 + 1/2 sin2x) = (1/2 + 1/2) sin2x - 1/2 = sin2x - 1/2.Wow, okay. So f(x) simplifies to sin2x - 1/2. That makes it much easier to work with.So for part (1), we need to find the monotonic intervals of f(x) = sin2x - 1/2.To find monotonic intervals, we need to find where the derivative is positive or negative.First, compute the derivative f’(x):f’(x) = 2 cos2x.So the derivative is 2 cos2x. The sign of f’(x) depends on cos2x.So, when is cos2x positive? When 2x is in (-π/2 + 2πk, π/2 + 2πk) for integers k, which means x is in (-π/4 + πk, π/4 + πk).Similarly, cos2x is negative when 2x is in (π/2 + 2πk, 3π/2 + 2πk), so x is in (π/4 + πk, 3π/4 + πk).Therefore, the function f(x) is increasing when cos2x > 0, i.e., when x is in (-π/4 + πk, π/4 + πk) for integers k.And decreasing when x is in (π/4 + πk, 3π/4 + πk) for integers k.Therefore, the monotonic intervals are intervals of π/2 length, alternating between increasing and decreasing.But since the question just says "find the monotonic intervals", they might be expecting intervals over a period, or in general. Let's check the original function's domain. The problem doesn't specify a domain, so we assume it's defined for all real numbers. Therefore, the monotonic intervals are all intervals of the form (-π/4 + πk, π/4 + πk) where the function is increasing, and (π/4 + πk, 3π/4 + πk) where it's decreasing, for all integers k.But maybe the problem expects the answer over a specific interval? Wait, no, the problem says "monotonic intervals", which typically refer to intervals where the function is increasing or decreasing. So since the function is periodic with period π (since sin2x has period π), the function f(x) has period π.Therefore, in each interval of length π, the function increases from -π/4 to π/4 and decreases from π/4 to 3π/4, and so on. So the monotonic intervals can be described as increasing on (-π/4 + kπ, π/4 + kπ) and decreasing on (π/4 + kπ, 3π/4 + kπ) for all integers k.But since the problem didn't specify a particular interval, this should be the answer for part (1).Now moving to part (2). In an acute triangle ABC, sides opposite angles A, B, C are a, b, c. Given that f(A/2) = (√3 -1)/2, and a=1, find the maximum value of the perimeter of triangle ABC.First, since f(x) = sin2x - 1/2, then f(A/2) = sin(2*(A/2)) - 1/2 = sinA - 1/2. So f(A/2) = sinA - 1/2 = (√3 -1)/2.Therefore, sinA - 1/2 = (√3 -1)/2. Let's solve for sinA.Adding 1/2 to both sides:sinA = (√3 -1)/2 + 1/2 = (√3 -1 +1)/2 = √3 / 2.Therefore, sinA = √3 / 2. So angle A is either π/3 or 2π/3. But since the triangle is acute, all angles are less than π/2. Wait, but wait, in an acute triangle, all angles are less than π/2 (90 degrees). Wait, but if sinA = √3/2, then A is π/3 (60 degrees), which is less than π/2. So A = π/3.Wait, hold on. If sinA = √3/2, the solutions in [0, π] are π/3 and 2π/3. But in an acute triangle, all angles must be less than π/2, so A = π/3 (60 degrees) is okay, but 2π/3 (120 degrees) would make the triangle obtuse. Therefore, since ABC is acute, A must be π/3.Therefore, angle A is π/3 radians, 60 degrees.Given that a=1, which is the side opposite angle A. So using the Law of Sines, we can relate the sides to the angles.Law of Sines: a / sinA = b / sinB = c / sinC = 2R, where R is the circumradius.Given a=1, sinA=√3/2, so 1 / (√3/2) = 2 / √3. Therefore, 2R = 2 / √3 => R = 1 / √3.Therefore, sides b = 2R sinB = (2 / √3) sinB, and c = (2 / √3) sinC.Perimeter P = a + b + c = 1 + (2 / √3)(sinB + sinC).Since the triangle is acute, all angles B and C must be less than π/2. Also, since A = π/3, angles B and C satisfy B + C = π - A = 2π/3. So C = 2π/3 - B.Therefore, perimeter P = 1 + (2 / √3)[sinB + sin(2π/3 - B)].Let me compute sin(2π/3 - B). Using sine subtraction formula:sin(2π/3 - B) = sin(2π/3)cosB - cos(2π/3)sinB.We know sin(2π/3) = √3/2 and cos(2π/3) = -1/2.Therefore, sin(2π/3 - B) = (√3/2)cosB - (-1/2)sinB = (√3/2)cosB + (1/2)sinB.Therefore, sinB + sin(2π/3 - B) = sinB + (√3/2 cosB + 1/2 sinB) = (sinB + 1/2 sinB) + (√3/2 cosB) = (3/2 sinB) + (√3/2 cosB).Therefore, perimeter P = 1 + (2 / √3)*(3/2 sinB + √3/2 cosB) = 1 + (2 / √3)*( (3 sinB + √3 cosB)/2 ) = 1 + ( (3 sinB + √3 cosB) / √3 ).Simplify the expression:(3 sinB + √3 cosB)/√3 = (3 / √3) sinB + (√3 / √3) cosB = √3 sinB + cosB.Therefore, perimeter P = 1 + √3 sinB + cosB.So now, we need to maximize P = 1 + √3 sinB + cosB, where angle B is such that the triangle is acute. Since the triangle is acute, angles B and C must be less than π/2. Given that A = π/3, and B + C = 2π/3, then both B and C must be in (0, π/2). Therefore, B ∈ (0, π/2) and C = 2π/3 - B ∈ (0, π/2). Let's find the range for B.Since C = 2π/3 - B < π/2, which implies 2π/3 - B < π/2 => B > 2π/3 - π/2 = π/6. Similarly, since B < π/2, and C = 2π/3 - B > 0, which implies 2π/3 - B > 0 => B < 2π/3. But since B must also be less than π/2, the upper bound is π/2. Therefore, B ∈ (π/6, π/2).Therefore, the domain for B is π/6 < B < π/2. So we need to maximize P = 1 + √3 sinB + cosB over B ∈ (π/6, π/2).To find the maximum, we can take the derivative of P with respect to B and set it to zero.Compute dP/dB: derivative of 1 is 0, derivative of √3 sinB is √3 cosB, derivative of cosB is -sinB. Therefore:dP/dB = √3 cosB - sinB.Set derivative to zero:√3 cosB - sinB = 0 => √3 cosB = sinB => tanB = √3.Therefore, tanB = √3 => B = π/3.Check if B = π/3 is within the interval (π/6, π/2). Yes, π/3 ≈ 1.047, which is between π/6 ≈ 0.523 and π/2 ≈ 1.570. So critical point at B = π/3.Now check if this is a maximum. Since the function is differentiable and the interval is open, the maximum must occur either at critical points or at endpoints. But since endpoints are excluded (B approaches π/6 or π/2), we check the critical point.Compute the second derivative to confirm concavity, but maybe we can just test values around π/3.Alternatively, since the function P is smooth, and the critical point is the only one in the interval, and as B approaches π/6 from the right:When B approaches π/6 (30 degrees), then C approaches 2π/3 - π/6 = π/2 (90 degrees), but since the triangle must be acute, C must be less than π/2. So approaching π/6, C approaches π/2, but stays less. Similarly, as B approaches π/2, C approaches 2π/3 - π/2 = π/6.Compute P at B = π/3:P = 1 + √3 sin(π/3) + cos(π/3) = 1 + √3*(√3/2) + 1/2 = 1 + (3/2) + 1/2 = 1 + 2 = 3.At B approaching π/6:P approaches 1 + √3 sin(π/6) + cos(π/6) = 1 + √3*(1/2) + (√3/2) = 1 + (√3/2 + √3/2) = 1 + √3 ≈ 1 + 1.732 ≈ 2.732.At B approaching π/2:P approaches 1 + √3 sin(π/2) + cos(π/2) = 1 + √3*1 + 0 = 1 + √3 ≈ 2.732.Therefore, the maximum perimeter is 3 when B = π/3. Therefore, the maximum perimeter is 3.But wait, let me check if when B = π/3, the triangle is still acute. If B = π/3, then C = 2π/3 - π/3 = π/3. So all angles are π/3, which is 60 degrees, which is acute. So the triangle is equilateral in this case. But given that a=1, then all sides are 1. Wait, but according to the Law of Sines, if all angles are π/3, then all sides are equal. Wait, but in our case, a=1. So if it's equilateral, all sides would be 1, perimeter 3. That checks out.Wait, but hold on: in our Law of Sines, we had 2R = 2/√3, so R = 1/√3. Then if all angles are π/3, each side would be 2R sin(π/3) = 2*(1/√3)*(√3/2) = 1. So sides a, b, c are all 1. Therefore, perimeter 3. So that is correct.But in the problem, it's an acute triangle. Since equilateral triangles are acute, this is valid.Therefore, the maximum perimeter is 3.But let me just confirm that there isn't a higher value somewhere else. But since at the critical point B=π/3, the perimeter is 3, and approaching the endpoints it's about 2.732, which is less than 3. Therefore, the maximum is indeed 3.Therefore, the answer to part (2) is 3.Final Answer(1) The function f(x) is increasing on intervals left( -frac{pi}{4} + kpi, frac{pi}{4} + kpi right) and decreasing on intervals left( frac{pi}{4} + kpi, frac{3pi}{4} + kpi right) for all integers k.(2) The maximum value of the perimeter of triangle ABC is boxed{3}.</think>Given the function ( f(x) = sin left( x + frac{5pi}{2} right) cos left( x - frac{pi}{2} right) - cos^2 left( x + frac{pi}{4} right) ):1. Monotonic Intervals of ( f(x) ): - Simplify the function using trigonometric identities: [ sin left( x + frac{5pi}{2} right) = cos(x), quad cos left( x - frac{pi}{2} right) = sin(x), quad cos^2 left( x + frac{pi}{4} right) = frac{1 - sin(2x)}{2} ] - The function simplifies to: [ f(x) = sin(2x) - frac{1}{2} ] - Compute the derivative: [ f'(x) = 2cos(2x) ] - Determine the intervals where ( f'(x) ) is positive or negative: - Increasing intervals: ( left( -frac{pi}{4} + kpi, frac{pi}{4} + kpi right) ) - Decreasing intervals: ( left( frac{pi}{4} + kpi, frac{3pi}{4} + kpi right) ) for all integers ( k ).2. Maximum Perimeter of ( triangle ABC ): - Given ( fleft( frac{A}{2} right) = frac{sqrt{3} - 1}{2} ): [ sin(A) - frac{1}{2} = frac{sqrt{3} - 1}{2} implies sin(A) = frac{sqrt{3}}{2} implies A = frac{pi}{3} ] - Using the Law of Sines with ( a = 1 ): [ frac{a}{sin(A)} = frac{2}{sqrt{3}} implies b = frac{2}{sqrt{3}} sin(B), quad c = frac{2}{sqrt{3}} sin(C) ] - Perimeter ( P ): [ P = 1 + frac{2}{sqrt{3}} (sin(B) + sin(C)) quad text{with} quad B + C = frac{2pi}{3} ] - Convert to a function of ( B ): [ P = 1 + sqrt{3} sin(B) + cos(B) ] - Maximize ( P ) for ( B in left( frac{pi}{6}, frac{pi}{2} right) ): - Critical point at ( B = frac{pi}{3} ) - Maximum perimeter: ( P = 3 ) Final Answer(1) The function ( f(x) ) is increasing on intervals ( left( -frac{pi}{4} + kpi, frac{pi}{4} + kpi right) ) and decreasing on intervals ( left( frac{pi}{4} + kpi, frac{3pi}{4} + kpi right) ) for all integers ( k ).(2) The maximum value of the perimeter of ( triangle ABC ) is (boxed{3}).]

🔑:(1) From the given function, we have f(x)=frac{sin 2x}{2}-frac{1+cos (2x+ frac{π}{2})}{2}=frac{sin 2x}{2}-frac{1-sin 2x}{2}=sin 2x-frac{1}{2}.Since - frac{π}{2}+2kπleqslant 2xleqslant frac{π}{2}+2kπ, kinmathbb{Z}, we have - frac{π}{4}+kπleqslant xleqslant frac{π}{4}+kπ, kinmathbb{Z}.And from frac{π}{2}+2kπleqslant 2xleqslant frac{3π}{2}+2kπ, kinmathbb{Z}, we have frac{π}{4}+kπleqslant xleqslant frac{3π}{4}+kπ, kinmathbb{Z}.Thus, the function f(x) is increasing on the intervals left[- frac{π}{4}+kπ, frac{π}{4}+kπright], kinmathbb{Z},and decreasing on the intervals left[frac{π}{4}+kπ, frac{3π}{4}+kπright], kinmathbb{Z}.(2) Given f(frac{A}{2})=sin A-frac{1}{2}=frac{sqrt{3}-1}{2}, we have sin A=frac{sqrt{3}}{2}. Since A is an acute angle, A=frac{π}{3}.With a=1, using the sine law, we get frac{b}{sin B}=frac{c}{sin C}=frac{a}{sin A}=frac{1}{frac{sqrt{3}}{2}}=frac{2sqrt{3}}{3}.Hence, b=frac{2sqrt{3}}{3}sin B, c=frac{2sqrt{3}}{3}sin C,and the perimeter of triangle ABC is L=a+b+c=1+frac{2sqrt{3}}{3}sin B+frac{2sqrt{3}}{3}sin C=1+frac{2sqrt{3}}{3}sin(frac{2π}{3}-C)+frac{2sqrt{3}}{3}sin C=1+2sin(C+frac{π}{6}).Thus, when sin(C+frac{π}{6}) takes its maximum value 1, the perimeter of triangle ABC attains its maximum value, which is boxed{3}.

🧠:[<think>Okay, so I need to find the value of 'a' such that the two given lines are perpendicular. The lines are x + 2y + 1 = 0 and ax + y - 2 = 0. The options are A: -2, B: -2/3, C: -1/3, D:1. Hmm, let me recall how to determine if two lines are perpendicular.First, I think the key is to look at their slopes. If two lines are perpendicular, the product of their slopes should be -1. Right? So I need to find the slopes of both lines and set their product equal to -1, then solve for 'a'.Let me start by converting both equations into slope-intercept form (y = mx + b), where m is the slope. Starting with the first line: x + 2y + 1 = 0. Let me solve for y.Subtract x and 1 from both sides: 2y = -x -1. Then divide by 2: y = (-1/2)x - 1/2. So the slope (m1) of the first line is -1/2.Now the second line: ax + y - 2 = 0. Solving for y. Subtract ax and add 2 to both sides: y = -ax + 2. So the slope (m2) of the second line is -a.Wait, if the product of the slopes should be -1 for perpendicular lines, then m1 * m2 = -1. Let me check that. So (-1/2) * (-a) = -1. Let me compute that.Multiply the two slopes: (-1/2) * (-a) = (a)/2. And this should equal -1. So (a)/2 = -1. Solving for a: multiply both sides by 2, so a = -2. Hmm, so that would be option A: -2. Let me confirm that.Wait, let me verify my steps again. First line: x + 2y +1 =0. Solving for y: 2y = -x -1, y = -1/2 x -1/2. Correct. So slope is -1/2. Second line: ax + y -2 =0. Solving for y: y = -ax +2. So slope is -a. Then their product is (-1/2)*(-a) = a/2. Set equal to -1: a/2 = -1. So a= -2. Yes, that seems correct. So answer A: -2.Wait, but let me check if I didn't mix up any signs. The second line is ax + y -2 =0, so y = -ax +2. So slope is -a. Then first slope is -1/2. Multiply them: (-1/2)*(-a) = (a)/2. So if the product is -1, then (a)/2 = -1 => a= -2. Correct.Alternatively, maybe there's another way to check. Let's think in terms of vectors. The normal vector of the first line is (1, 2), since the equation is x + 2y +1 =0. The normal vector of the second line is (a, 1). For two lines to be perpendicular, their normal vectors must also be perpendicular. Wait, is that true?Wait, actually, the normal vector is perpendicular to the line itself. So if two lines are perpendicular, their normal vectors would be parallel? Wait, maybe I confused direction vectors. Let me think again.No, maybe it's better to stick with the slopes. Because the direction vector of a line is related to its slope. For example, a line with slope m has a direction vector of (1, m). Then, for two lines to be perpendicular, the dot product of their direction vectors should be zero.So first line's slope is -1/2, so direction vector is (1, -1/2). Second line's slope is -a, so direction vector is (1, -a). The dot product should be zero. Let's compute that:(1)(1) + (-1/2)(-a) = 1 + (a)/2. Set equal to zero: 1 + a/2 =0. Then a/2 = -1 => a= -2. Same result. So that's consistent.Alternatively, if we consider normal vectors: the normal vector to the first line is (1, 2), and the normal vector to the second line is (a, 1). Wait, but for the lines to be perpendicular, the normal vectors need to satisfy some condition? Maybe not directly. Because the normal vector is perpendicular to the line, so if two lines are perpendicular, their normal vectors would form some angle. Let me check.If line L1 has normal vector N1, and line L2 has normal vector N2, then the angle between N1 and N2 would be equal to the angle between the lines themselves? Hmm, not sure. Maybe it's better to stick with direction vectors or slopes.But anyway, both methods using slopes give a = -2. So the answer should be A. But let me check with plugging in a = -2 into the second line's equation. Then the second line becomes -2x + y -2 =0, or y = 2x +2. So its slope is 2. The first line has slope -1/2. Then 2 * (-1/2) = -1. Yes, that's correct, their product is -1. So they are perpendicular. Therefore, a = -2 is indeed the correct answer. So the answer is A.Final Answerboxed{A}</think>To determine the value of ( a ) such that the lines ( x + 2y + 1 = 0 ) and ( ax + y - 2 = 0 ) are perpendicular, we need to find their slopes and ensure their product is (-1).1. Convert the first line ( x + 2y + 1 = 0 ) to slope-intercept form: [ x + 2y + 1 = 0 implies 2y = -x - 1 implies y = -frac{1}{2}x - frac{1}{2} ] The slope (( m_1 )) of this line is ( -frac{1}{2} ).2. Convert the second line ( ax + y - 2 = 0 ) to slope-intercept form: [ ax + y - 2 = 0 implies y = -ax + 2 ] The slope (( m_2 )) of this line is ( -a ).3. For the lines to be perpendicular, the product of their slopes must be (-1): [ m_1 cdot m_2 = -1 implies left( -frac{1}{2} right) cdot (-a) = -1 ] Simplifying this equation: [ frac{a}{2} = -1 implies a = -2 ]4. Verify by checking the product of the slopes when ( a = -2 ): - The second line becomes ( -2x + y - 2 = 0 ) or ( y = 2x + 2 ), with a slope of ( 2 ). - The product of the slopes ( 2 ) and ( -frac{1}{2} ) is ( 2 cdot -frac{1}{2} = -1 ), confirming they are perpendicular.Thus, the value of ( a ) is (boxed{A}).,<think>Okay, let's see. I need to find the value of 'a' such that the two given lines are perpendicular. The lines are x + 2y + 1 = 0 and ax + y - 2 = 0. The options are A: -2, B: -2/3, C: -1/3, D: 1. Hmm, right. So perpendicular lines have slopes that are negative reciprocals of each other. That is, if one line has a slope of m, the other should have a slope of -1/m. So first, I need to find the slopes of these two lines.Let me start with the first line: x + 2y + 1 = 0. To find its slope, I can convert it into slope-intercept form (y = mx + b). So, solving for y:x + 2y + 1 = 0 Subtract x and 1 from both sides: 2y = -x - 1 Divide by 2: y = (-1/2)x - 1/2So the slope (m1) of the first line is -1/2.Now, the second line is ax + y - 2 = 0. Let's convert that to slope-intercept form too.ax + y - 2 = 0 Subtract ax and add 2 to both sides: y = -ax + 2So the slope (m2) of the second line is -a.Since the lines are perpendicular, the product of their slopes should be -1. So:m1 * m2 = -1 Substituting the slopes we found: (-1/2) * (-a) = -1Let me compute that:(-1/2) * (-a) = (1/2)a So (1/2)a = -1 Multiply both sides by 2: a = -2Wait, so a is -2? But let me check the options. Option A is -2. So that's the answer? Let me verify again to make sure I didn't make a mistake.First line: slope -1/2. Second line: slope -a. For perpendicularity, their product should be -1. So (-1/2)*(-a) = (a)/2. Setting that equal to -1: a/2 = -1 => a = -2. Yes, that seems correct.Alternatively, maybe I can check by plugging in a = -2 into the second equation and seeing if they are perpendicular. If a is -2, the second line becomes -2x + y - 2 = 0, which is y = 2x + 2. Its slope is 2. The first line's slope is -1/2. Multiplying them: 2 * (-1/2) = -1, which confirms they are perpendicular. So yes, a = -2 is correct. The answer should be A.Final Answerboxed{A}</think>To determine the value of (a) such that the lines (x + 2y + 1 = 0) and (ax + y - 2 = 0) are perpendicular, we need to find the slopes of these lines and ensure they are negative reciprocals of each other.First, convert the line (x + 2y + 1 = 0) to slope-intercept form:[x + 2y + 1 = 0 implies 2y = -x - 1 implies y = -frac{1}{2}x - frac{1}{2}]The slope ((m_1)) of this line is (-frac{1}{2}).Next, convert the line (ax + y - 2 = 0) to slope-intercept form:[ax + y - 2 = 0 implies y = -ax + 2]The slope ((m_2)) of this line is (-a).For the lines to be perpendicular, the product of their slopes must be (-1):[m_1 cdot m_2 = -1 implies left(-frac{1}{2}right) cdot (-a) = -1]Simplifying this equation:[frac{a}{2} = -1 implies a = -2]Verification: If (a = -2), the second line becomes (y = 2x + 2) with a slope of (2). The product of the slopes (-frac{1}{2}) and (2) is (-1), confirming perpendicularity.Thus, the value of (a) is (boxed{A}).]

🔑:Since the slope of the line x+2y+1=0 exists, and the line x+2y+1=0 is perpendicular to the line ax+y-2=0, then - frac{1}{2} times (-a) = -1, solving this gives a = -2. Therefore, the answer is: boxed{text{A}}. This can be determined by using the relationship between the slopes of two perpendicular lines. This question tests the relationship between the slopes of two perpendicular lines, as well as reasoning and computational skills, and is considered a basic question.

🧠:[<think>Alright, let's try to figure out this problem. So, we need to find the sum of the real numbers x that satisfy the equation:(2x)/(x² + 5x + 3) + (3x)/(x² + x + 3) = 1Hmm, okay. Let me start by understanding what's going on here. The equation has two rational expressions added together equal to 1. My goal is to find all real solutions x and then sum them up. The question is specifically asking for the sum, so maybe there's a clever way to find the sum without finding each root individually. But I might need to solve the equation first.First, let me note that since there are denominators, I need to make sure that the denominators are not zero. So, the solutions x must not make x² + 5x + 3 = 0 or x² + x + 3 = 0. Let me check the discriminants of these quadratics to see if there are any real roots. For x² + 5x + 3, discriminant D1 = 25 - 12 = 13, which is positive, so there are two real roots. For x² + x + 3, discriminant D2 = 1 - 12 = -11, which is negative, so no real roots. That means the denominator x² + x + 3 is never zero for real x, so the only restrictions on x come from x² + 5x + 3 ≠ 0. But maybe we can handle that later.Now, moving on to solving the equation. Let's denote the equation again:(2x)/(x² + 5x + 3) + (3x)/(x² + x + 3) = 1To solve this, I might need to combine the fractions. To do that, I need a common denominator, which would be (x² + 5x + 3)(x² + x + 3). That seems a bit complicated, but let's try.Multiply both sides by the common denominator:2x(x² + x + 3) + 3x(x² + 5x + 3) = (x² + 5x + 3)(x² + x + 3)Okay, let's expand each term step by step.First term: 2x(x² + x + 3) = 2x³ + 2x² + 6xSecond term: 3x(x² + 5x + 3) = 3x³ + 15x² + 9xLeft-hand side (LHS) total: 2x³ + 2x² + 6x + 3x³ + 15x² + 9x = (2x³ + 3x³) + (2x² + 15x²) + (6x + 9x) = 5x³ + 17x² + 15xRight-hand side (RHS): (x² + 5x + 3)(x² + x + 3)Let me expand that. Multiply each term:First, x²*(x² + x + 3) = x⁴ + x³ + 3x²Then, 5x*(x² + x + 3) = 5x³ + 5x² + 15xThen, 3*(x² + x + 3) = 3x² + 3x + 9So, adding all those together:x⁴ + x³ + 3x² + 5x³ + 5x² + 15x + 3x² + 3x + 9Combine like terms:x⁴ + (1x³ + 5x³) + (3x² + 5x² + 3x²) + (15x + 3x) + 9Which is:x⁴ + 6x³ + 11x² + 18x + 9So, RHS is x⁴ + 6x³ + 11x² + 18x + 9Therefore, bringing everything to one side, we get:LHS - RHS = 0So:5x³ + 17x² + 15x - (x⁴ + 6x³ + 11x² + 18x + 9) = 0Let's distribute the negative sign:5x³ + 17x² + 15x - x⁴ - 6x³ - 11x² - 18x - 9 = 0Now, combine like terms:- x⁴ + (5x³ - 6x³) + (17x² - 11x²) + (15x - 18x) - 9 = 0Calculating each:- x⁴ - x³ + 6x² - 3x - 9 = 0Multiply both sides by -1 to make the leading coefficient positive:x⁴ + x³ - 6x² + 3x + 9 = 0Hmm, okay, so now we have a quartic equation: x⁴ + x³ - 6x² + 3x + 9 = 0Quartic equations can be tough, but maybe this factors. Let's try to factor it. Let's check for rational roots using Rational Root Theorem. Possible rational roots are ±1, ±3, ±9.Let's test x = 1:1 + 1 - 6 + 3 + 9 = 1 +1 -6 +3 +9 = 8. Not zero.x = -1:1 -1 -6 -3 +9 = 0. Wait, 1 -1 = 0, 0 -6 = -6, -6 -3 = -9, -9 +9 = 0. So x = -1 is a root.So, (x + 1) is a factor. Let's perform polynomial division.Dividing x⁴ + x³ - 6x² + 3x + 9 by (x + 1). Let's use synthetic division.Set up coefficients: 1 (x⁴), 1 (x³), -6 (x²), 3 (x), 9 (constant)Root: x = -1Bring down the 1.Multiply by -1: 1*(-1) = -1. Add to next coefficient: 1 + (-1) = 0.Multiply by -1: 0*(-1) = 0. Add to next coefficient: -6 + 0 = -6.Multiply by -1: -6*(-1) = 6. Add to next coefficient: 3 +6=9.Multiply by -1:9*(-1)= -9. Add to last coefficient:9 + (-9)=0. So, remainder is zero.So the quotient is x³ + 0x² -6x +9, so x³ -6x +9.Therefore, the quartic factors as (x +1)(x³ -6x +9). Now, let's factor the cubic x³ -6x +9.Again, try Rational Root Theorem: possible roots ±1, ±3, ±9.Test x =1:1 -6 +9=4≠0x= -1: -1 +6 +9=14≠0x=3:27 -18 +9=18≠0x=-3:-27 +18 +9=0. Wait, -27 +18= -9 +9=0. So x= -3 is a root.Therefore, (x +3) is a factor. Let's divide x³ -6x +9 by (x +3).Using synthetic division:Coefficients: 1 (x³), 0 (x²), -6 (x), 9 (constant)Root: x=-3Bring down 1.Multiply by -3:1*(-3)=-3. Add to next coefficient:0 + (-3)=-3.Multiply by -3:-3*(-3)=9. Add to next coefficient: -6 +9=3.Multiply by -3:3*(-3)=-9. Add to last coefficient:9 + (-9)=0.So quotient is x² -3x +3.Thus, the cubic factors as (x +3)(x² -3x +3). Therefore, the quartic factors as (x +1)(x +3)(x² -3x +3).Therefore, the equation is (x +1)(x +3)(x² -3x +3)=0.So, the roots are x = -1, x = -3, and the roots of x² -3x +3=0. Let's check the discriminant of x² -3x +3: D=9 -12 = -3 <0, so no real roots. Therefore, the only real solutions are x=-1 and x=-3.Therefore, the sum of the real numbers x is (-1) + (-3) = -4.But wait, let me verify this because sometimes when we manipulate equations, especially multiplying both sides by denominators, we might introduce extraneous solutions. So, let's check if x=-1 and x=-3 actually satisfy the original equation.First, check x = -1:Left-hand side (LHS): [2*(-1)]/[(-1)^2 +5*(-1) +3] + [3*(-1)]/[(-1)^2 + (-1) +3]Compute denominators:First denominator: 1 -5 +3 = -1Second denominator:1 -1 +3 = 3So, LHS: (-2)/(-1) + (-3)/3 = 2 + (-1) = 1. Which equals RHS. So x=-1 is valid.Check x = -3:First denominator: (-3)^2 +5*(-3) +3 =9 -15 +3= -3Second denominator: (-3)^2 + (-3) +3=9 -3 +3=9So LHS: [2*(-3)]/(-3) + [3*(-3)]/9 = (-6)/(-3) + (-9)/9 = 2 + (-1) =1. Which matches RHS. So x=-3 is valid too.Therefore, both solutions are valid. Since the other roots are complex, the only real solutions are x=-1 and x=-3. Their sum is -4.So, the answer is boxed{-4}.Final Answerboxed{-4}</think>To determine the sum of the real numbers ( x ) for which (frac{2 x}{x^{2}+5 x+3}+frac{3 x}{x^{2}+x+3}=1), we start by combining the fractions:[frac{2x}{x^2 + 5x + 3} + frac{3x}{x^2 + x + 3} = 1]We multiply both sides by the common denominator ((x^2 + 5x + 3)(x^2 + x + 3)):[2x(x^2 + x + 3) + 3x(x^2 + 5x + 3) = (x^2 + 5x + 3)(x^2 + x + 3)]Expanding and simplifying both sides, we get:Left-hand side (LHS):[2x(x^2 + x + 3) + 3x(x^2 + 5x + 3) = 5x^3 + 17x^2 + 15x]Right-hand side (RHS):[(x^2 + 5x + 3)(x^2 + x + 3) = x^4 + 6x^3 + 11x^2 + 18x + 9]Subtracting RHS from LHS:[5x^3 + 17x^2 + 15x - (x^4 + 6x^3 + 11x^2 + 18x + 9) = 0]Simplifying:[-x^4 - x^3 + 6x^2 - 3x - 9 = 0]Multiplying by -1:[x^4 + x^3 - 6x^2 + 3x + 9 = 0]Factoring the quartic polynomial:[(x + 1)(x + 3)(x^2 - 3x + 3) = 0]The quadratic (x^2 - 3x + 3) has no real roots (discriminant (-3)). The real roots are (x = -1) and (x = -3). Verifying these solutions in the original equation confirms they are valid.The sum of the real solutions is:[-1 + (-3) = -4]Thus, the answer is (boxed{-4}).,<think>Alright, let's tackle this problem step by step. The question is asking for the sum of the real numbers ( x ) that satisfy the equation:[frac{2x}{x^{2} + 5x + 3} + frac{3x}{x^{2} + x + 3} = 1]Hmm, okay. So, we need to find all real solutions ( x ) and then sum them up. The problem mentions "real numbers," so maybe there are multiple solutions, and after finding them, we just add them together. I remember that for quadratic equations, the sum of roots is ( -b/a ), but this equation isn't obviously a quadratic. Let's see how to approach this.First, since there are two fractions here, maybe I can combine them by finding a common denominator. The denominators are ( x^{2} + 5x + 3 ) and ( x^{2} + x + 3 ). Let me check if these denominators can be factored or if they have any common factors. Let's compute the discriminants to see if they can be factored.For ( x^{2} + 5x + 3 ):Discriminant ( D = 25 - 12 = 13 ), which is positive, so it has two real roots. Similarly, for ( x^{2} + x + 3 ):Discriminant ( D = 1 - 12 = -11 ), which is negative, so this quadratic doesn't factor over the reals.So, the denominators can't be factored easily, but at least one of them is always positive (since ( x^{2} + x + 3 ) has no real roots and the coefficient of ( x^2 ) is positive, it's always positive). The other denominator, ( x^{2} + 5x + 3 ), can be zero for some real ( x ), so we need to make sure we don't include those ( x ) values in our solutions.But maybe instead of factoring, we can cross-multiply or manipulate the equation to make it easier. Let's try to combine the two fractions:Let me denote ( A = x^{2} + 5x + 3 ) and ( B = x^{2} + x + 3 ). Then the equation becomes:[frac{2x}{A} + frac{3x}{B} = 1]Multiply both sides by ( AB ):[2x cdot B + 3x cdot A = AB]Expanding each term:First, ( 2x cdot B = 2x(x^{2} + x + 3) = 2x^3 + 2x^2 + 6x )Second, ( 3x cdot A = 3x(x^{2} + 5x + 3) = 3x^3 + 15x^2 + 9x )Third, ( AB = (x^{2} + 5x + 3)(x^{2} + x + 3) ). Let me compute this:Multiply term by term:First, ( x^2 times x^2 = x^4 )Then, ( x^2 times x = x^3 )Then, ( x^2 times 3 = 3x^2 )Next, ( 5x times x^2 = 5x^3 )Then, ( 5x times x = 5x^2 )Then, ( 5x times 3 = 15x )Next, ( 3 times x^2 = 3x^2 )Then, ( 3 times x = 3x )Then, ( 3 times 3 = 9 )Adding all these terms:( x^4 + (1x^3 + 5x^3) + (3x^2 + 5x^2 + 3x^2) + (15x + 3x) + 9 )Simplify:( x^4 + 6x^3 + 11x^2 + 18x + 9 )So, AB is ( x^4 + 6x^3 + 11x^2 + 18x + 9 )Now, going back to the equation:Left side after expanding: ( 2x^3 + 2x^2 + 6x + 3x^3 + 15x^2 + 9x )Combine like terms:( (2x^3 + 3x^3) = 5x^3 )( (2x^2 + 15x^2) = 17x^2 )( (6x + 9x) = 15x )So, left side is ( 5x^3 + 17x^2 + 15x )The equation is:( 5x^3 + 17x^2 + 15x = x^4 + 6x^3 + 11x^2 + 18x + 9 )Bring all terms to the right side:( 0 = x^4 + 6x^3 + 11x^2 + 18x + 9 - 5x^3 -17x^2 -15x )Simplify:Combine ( x^4 ): ( x^4 )Combine ( x^3 ): ( 6x^3 -5x^3 = x^3 )Combine ( x^2 ): ( 11x^2 -17x^2 = -6x^2 )Combine ( x ): ( 18x -15x = 3x )Constants: ( +9 )So, the equation becomes:( x^4 + x^3 -6x^2 + 3x + 9 = 0 )Hmm, so we have a quartic equation. Quartic equations can be tricky, but maybe this factors into lower-degree polynomials. Let's try to factor it.First, let's check for rational roots using Rational Root Theorem. Possible rational roots are factors of 9 divided by factors of 1, so possible candidates: ±1, ±3, ±9.Let's test x = 1:( 1 +1 -6 +3 +9 = 1 +1 -6 +3 +9 = 8 ≠ 0 )x = -1:( 1 -1 -6 -3 +9 = 0. Let's compute: 1 -1 = 0; 0 -6 = -6; -6 -3 = -9; -9 +9 = 0. So, x = -1 is a root.Therefore, (x +1) is a factor. Let's perform polynomial division or use synthetic division.Using synthetic division for root x = -1:Coefficients: 1 (x^4), 1 (x^3), -6 (x^2), 3 (x), 9 (constant)Set up synthetic division:-1 | 1   1   -6   3   9Start: Bring down the 1.Multiply by -1: 1*(-1) = -1. Add to next coefficient: 1 + (-1) = 0.Multiply by -1: 0*(-1) = 0. Add to next coefficient: -6 +0 = -6.Multiply by -1: -6*(-1) =6. Add to next coefficient: 3 +6 =9.Multiply by -1:9*(-1) = -9. Add to last coefficient:9 + (-9)=0.So, the result is coefficients: 1   0   -6   9, which corresponds to:( x^3 +0x^2 -6x +9 )So, the quartic factors as (x +1)(x^3 -6x +9). Now, let's factor the cubic ( x^3 -6x +9 ).Again, try rational roots: possible roots are ±1, ±3, ±9.Test x=1:1 -6 +9 =4 ≠0x= -1:-1 +6 +9=14≠0x=3:27 -18 +9=18≠0x=-3:-27 +18 +9=0. Wait, x=-3:-27 -6*(-3) +9 = -27 +18 +9=0. Wait, let's compute properly:x=-3:(-3)^3 -6*(-3) +9 = -27 +18 +9 =0. Yes, x=-3 is a root.Therefore, (x +3) is a factor. Let's perform synthetic division on the cubic:Divide ( x^3 -6x +9 ) by (x +3).Set up synthetic division for x=-3:-3 | 1   0   -6   9Bring down 1.Multiply by -3: 1*(-3) = -3. Add to next coefficient: 0 + (-3) = -3.Multiply by -3: -3*(-3)=9. Add to next coefficient: -6 +9=3.Multiply by -3:3*(-3)=-9. Add to next coefficient:9 + (-9)=0.So, the cubic factors as (x +3)(x^2 -3x +3). Therefore, the quartic factors as:(x +1)(x +3)(x^2 -3x +3)Now, set each factor equal to zero:x +1 =0 ⇒ x= -1x +3 =0 ⇒x= -3x^2 -3x +3=0 ⇒ discriminant D=9 -12= -3 <0, so no real roots.Therefore, the real solutions are x= -1 and x= -3. Hence, the sum is (-1) + (-3)= -4.Wait, but hold on. Let me verify this. The original equation is:[frac{2x}{x^{2} + 5x + 3} + frac{3x}{x^{2} + x + 3} = 1]We found roots x= -1 and x= -3, but we need to make sure that these x values don't make any denominators zero. Let's check:For x= -1:First denominator: (-1)^2 +5*(-1)+3=1 -5 +3= -1 ≠0Second denominator: (-1)^2 +(-1)+3=1 -1 +3=3 ≠0. Okay, valid.For x= -3:First denominator: (-3)^2 +5*(-3)+3=9 -15 +3= -3. Wait, -3. So denominator is -3. But denominator can't be zero. Since -3 ≠0, okay.Wait, x= -3: denominator1 is -3, which is okay, as long as not zero. So both x= -1 and x= -3 are valid solutions.Therefore, the sum is -1 + (-3) = -4.But wait, is that all? Let me check if there are more solutions. The quartic equation had two real roots and two complex roots. So, the only real solutions are x=-1 and x=-3. Therefore, their sum is -4. So, the answer should be boxed{-4}But just to make sure, let's plug x=-1 into the original equation:First term: 2*(-1)/[(-1)^2 +5*(-1)+3] = -2/(1 -5 +3)= -2/(-1)=2Second term:3*(-1)/[(-1)^2 + (-1) +3] = -3/(1 -1 +3)= -3/3= -1So total:2 + (-1)=1. Correct.For x= -3:First term:2*(-3)/[9 +5*(-3)+3] = -6/(9 -15 +3)= -6/(-3)=2Second term:3*(-3)/[9 + (-3)+3]= -9/(9 -3 +3)= -9/9= -1Total:2 + (-1)=1. Correct.So both solutions work, and there are no other real solutions. Thus, the sum is indeed -4.Final Answerboxed{-4}</think>To determine the sum of the real numbers ( x ) for which (frac{2 x}{x^{2}+5 x+3}+frac{3 x}{x^{2}+x+3}=1), we start by combining the fractions:[frac{2x}{x^2 + 5x + 3} + frac{3x}{x^2 + x + 3} = 1]Multiply both sides by ((x^2 + 5x + 3)(x^2 + x + 3)) to clear the denominators:[2x(x^2 + x + 3) + 3x(x^2 + 5x + 3) = (x^2 + 5x + 3)(x^2 + x + 3)]Expanding and simplifying both sides, we get:Left side:[2x(x^2 + x + 3) + 3x(x^2 + 5x + 3) = 2x^3 + 2x^2 + 6x + 3x^3 + 15x^2 + 9x = 5x^3 + 17x^2 + 15x]Right side:[(x^2 + 5x + 3)(x^2 + x + 3) = x^4 + 6x^3 + 11x^2 + 18x + 9]Equating both sides and simplifying:[5x^3 + 17x^2 + 15x = x^4 + 6x^3 + 11x^2 + 18x + 9][x^4 + x^3 - 6x^2 + 3x + 9 = 0]Using the Rational Root Theorem and synthetic division, we factor the quartic equation:[(x + 1)(x + 3)(x^2 - 3x + 3) = 0]The quadratic (x^2 - 3x + 3) has no real roots (discriminant (-3)). The real solutions are (x = -1) and (x = -3). Verifying these solutions in the original equation confirms they are valid. The sum of the real solutions is:[-1 + (-3) = -4]Thus, the final answer is (boxed{-4}).]

🔑:By clearing denominators, the equation becomes2 xleft(x^{2}+x+3right)+3 xleft(x^{2}+5 x+3right)=left(x^{2}+5 x+3right)left(x^{2}+x+3right)and after expanding the left side, we get2 x^{3}+2 x^{2}+6 x+3 x^{3}+15 x^{2}+9 x=left(x^{2}+5 x+3right)left(x^{2}+x+3right) text {. }There are many ways to systematically expand the right side. One way is to think of it as a binomialbegin{aligned}left(x^{2}+5 x+3right)left(x^{2}+x+3right) & =left(left(x^{2}+(5 x+3)right)left(x^{2}+(x+3)right)right. & =x^{4}+x^{2}(x+3)+(5 x+3) x^{2}+(5 x+3)(x+3) & =x^{4}+x^{3}+3 x^{2}+5 x^{3}+3 x^{2}+5 x^{2}+15 x+3 x+9 & =x^{4}+6 x^{3}+11 x^{2}+18 x+9 .end{aligned}The original equation is equivalent to2 x^{3}+2 x^{2}+6 x+3 x^{3}+15 x^{2}+9 x=x^{4}+6 x^{3}+11 x^{2}+18 x+9which can be rearranged tox^{4}+x^{3}-6 x^{2}+3 x+9=0 text {. }Generally, finding the real roots of a quartic polynomial is quite difficult, but the rational roots theorem tells us that any rational roots of the quartic above must be divisors of 9. In general, there could be real roots that are not rational, but that will not be the case here.The possible rational roots of the quartic are pm 1, pm 3, and pm 9.Checking x=1, we find that (1)^{4}+(1)^{3}-6(1)^{2}+3(1)+9=8, so 1 is not a root.Checking x=-1, we find that (-1)^{4}+(-1)^{3}-6(-1)^{2}+3(-1)+9=1-1-6-3+9=0 so -1 is a root.This means x+1 is a factor of the quartic. After factoring, we get thatx^{4}+x^{3}-6 x^{2}+3 x+9=(x+1)left(x^{3}-6 x+9right) text {. }We now try to find a rational root of x^{3}-6 x+9. Again, the possibilities are pm 1, pm 3, and pm 9. Of these six possibilities, only -3 is a root:(-3)^{3}-6(-3)+9=-27+18+9=0 .This means x+3 mathrm{~s} a factor of x^{3}-6 x+9, and after factoring, we get thatx^{4}+x^{3}-6 x^{2}+3 x+9=(x+1)(x+3)left(x^{2}-3 x+3right) text {. }Thus, the equation we wish to solve is equivalent to(x+3)(x+1)left(x^{2}-3 x+3right)=0 text {. }The roots of the quartic are therefore x=-1 and x=-3 and the roots of x^{2}-3 x+3.Using the quadratic formula, the two roots of x^{2}-3 x+3 are frac{3+sqrt{-3}}{2} and frac{3-sqrt{-3}}{2}, neither of which are real.Therefore, the only real values of x that can possibly satisfy the original equation are -3 and -1 .It can be checked that both x=-1 and x=-3 satisfy the original equation, so the answer is (-1)+(-3)=-4.