🧠:[<think>Okay, so I need to solve the equation cos x cos 2x sin 3x = 0.25 sin 2x. Hmm, let me see. First, I should probably try to simplify both sides of the equation. Maybe using some trigonometric identities can help here. Let me recall some identities.I know that products of trigonometric functions can sometimes be converted into sums or differences using product-to-sum formulas. Let me write down the left side: cos x cos 2x sin 3x. That's a product of three functions. Maybe I can group cos x and cos 2x first?Yes, there's a product-to-sum identity for cos A cos B. The formula is cos A cos B = [cos(A+B) + cos(A-B)] / 2. Let me apply that to cos x cos 2x.So, cos x cos 2x = [cos(3x) + cos(-x)] / 2. But cos(-x) is the same as cos x, so this simplifies to [cos 3x + cos x] / 2. Okay, so the left side becomes [cos 3x + cos x]/2 multiplied by sin 3x. So, the entire left side is (cos 3x + cos x)/2 * sin 3x.Let me distribute the sin 3x: [cos 3x sin 3x + cos x sin 3x]/2. Now, maybe apply product-to-sum formulas again here. For the first term, cos 3x sin 3x. There's an identity for cos A sin B: [sin(A+B) + sin(B - A)] / 2. Wait, cos A sin B = [sin(A+B) + sin(B - A)] / 2. Let me check that.Yes, more accurately, the identity is cos A sin B = [sin(A + B) + sin(B - A)] / 2. So applying that to cos 3x sin 3x: A is 3x, B is 3x. So, sin(3x + 3x) + sin(3x - 3x) over 2. That simplifies to sin 6x + sin 0 over 2. But sin 0 is 0, so this term is just sin 6x / 2.Then, the second term is cos x sin 3x. Again, using the same identity: cos x sin 3x = [sin(x + 3x) + sin(3x - x)] / 2 = [sin 4x + sin 2x] / 2.Therefore, putting it all back together, the left side becomes [ (sin 6x / 2) + (sin 4x + sin 2x)/2 ] / 2. Wait, because the original expression after the first product-to-sum was divided by 2. Let me retrace:Original left side was cos x cos 2x sin 3x = [cos 3x + cos x]/2 * sin 3x. Then, distributing sin 3x gives [cos 3x sin 3x + cos x sin 3x]/2. Then, each of those terms is further transformed:cos 3x sin 3x = sin 6x / 2, and cos x sin 3x = [sin 4x + sin 2x]/2. So, substituting back, we have [ (sin 6x / 2) + (sin 4x + sin 2x)/2 ] / 2.Wait, so that's [(sin 6x + sin 4x + sin 2x)/2] divided by 2, which is (sin 6x + sin 4x + sin 2x)/4. So the left side is (sin 6x + sin 4x + sin 2x)/4. The original equation is then:(sin 6x + sin 4x + sin 2x)/4 = 0.25 sin 2x.But 0.25 is 1/4, so multiply both sides by 4 to eliminate denominators:sin 6x + sin 4x + sin 2x = sin 2x.Subtract sin 2x from both sides:sin 6x + sin 4x = 0.So now, the equation simplifies to sin 6x + sin 4x = 0. Hmm. Let's work on this.We can use the sum-to-product identity here. The identity for sin A + sin B is 2 sin[(A+B)/2] cos[(A-B)/2]. Let's apply that to sin 6x + sin 4x.So, A = 6x, B = 4x. Then:sin 6x + sin 4x = 2 sin[(6x + 4x)/2] cos[(6x - 4x)/2] = 2 sin(5x) cos x.So the equation becomes 2 sin 5x cos x = 0.Therefore, 2 sin 5x cos x = 0. Since 2 is not zero, we can divide both sides by 2:sin 5x cos x = 0.So, the product sin 5x cos x equals zero. Therefore, either sin 5x = 0 or cos x = 0.Let me solve each case separately.Case 1: sin 5x = 0.The solutions to sin θ = 0 are θ = nπ, where n is an integer. So, 5x = nπ => x = nπ/5.Case 2: cos x = 0.The solutions to cos θ = 0 are θ = π/2 + nπ, where n is an integer. So, x = π/2 + nπ.Therefore, the solutions are x = nπ/5 or x = π/2 + nπ, for any integer n.But we need to check if all these solutions are valid, because sometimes when we manipulate trigonometric equations, especially multiplying/dividing or using identities, we might introduce extraneous solutions or lose some solutions.Wait, let me check if all these solutions satisfy the original equation. Let's verify with some test values.First, let's check x = π/5. Let's compute left side and right side.Left side: cos(π/5) cos(2π/5) sin(3π/5).Right side: 0.25 sin(2π/5).Let me compute these numerically. π ≈ 3.1416, so π/5 ≈ 0.6283, 2π/5 ≈ 1.2566, 3π/5 ≈ 1.8849.cos(π/5) ≈ 0.8090, cos(2π/5) ≈ 0.3090, sin(3π/5) ≈ 0.5878.Left side: 0.8090 * 0.3090 * 0.5878 ≈ 0.8090 * 0.3090 ≈ 0.2499, then 0.2499 * 0.5878 ≈ 0.1468.Right side: 0.25 * sin(2π/5) ≈ 0.25 * 0.5878 ≈ 0.14695. So left ≈ 0.1468, right ≈ 0.14695. Close enough, considering rounding errors. So x = π/5 is a valid solution.Now, let's check x = π/2. Then cos x = 0, so left side is 0 * cos(2x) sin(3x) = 0. Right side is 0.25 sin(2*(π/2)) = 0.25 sin π = 0. So 0 = 0. Valid solution.Another solution: x = 0. Let's check x = 0. Left side: cos 0 * cos 0 * sin 0 = 1 * 1 * 0 = 0. Right side: 0.25 sin 0 = 0. So 0=0. Valid. x=0 is a solution. But according to our solution set, x = nπ/5. For n=0, x=0; n=1, x=π/5, etc. So x=0 is included. So maybe all solutions are valid.But let's check another solution from the second case: x = π/2 + π = 3π/2. Then, cos(3π/2) = 0. Left side is 0, right side: 0.25 sin(2*(3π/2)) = 0.25 sin(3π) = 0.25*0 = 0. So that's valid.Wait, but what about x = π/2. Let's check if sin 5x = 0 here. 5x = 5*(π/2) = 5π/2. sin(5π/2) = sin(π/2) = 1. So sin 5x is not zero here. So x=π/2 is covered under the cos x=0 case. Similarly, x=3π/2: cos x=0, and 5x=15π/2, sin(15π/2)=sin(7π + π/2)=sin(π/2)=1. So again, not zero. So the two cases are distinct. So no overlap. So all solutions are either in x = nπ/5 or x = π/2 + nπ.But wait, let's check if there is overlap. Suppose nπ/5 = π/2 + kπ, for integers n and k. Then, solving for n: n/5 = 1/2 + k => n = 5/2 +5k. But n must be an integer, so 5/2 +5k is not an integer unless k is a half-integer, which it isn't. Therefore, no overlapping solutions. Thus, the solution set is the union of both cases.Therefore, the general solution is x = nπ/5 or x = π/2 + nπ for any integer n.But let me check if there are any restrictions. For example, when we applied the product-to-sum identities, did we divide by any terms that could be zero? Let me see. The steps were:Original equation: cos x cos 2x sin 3x = 0.25 sin 2x.We transformed the left side using identities, leading us to sin 6x + sin 4x + sin 2x = sin 2x, then subtract sin 2x to get sin 6x + sin 4x = 0. Then converted to 2 sin5x cosx =0, leading to sin5x=0 or cosx=0.But in that process, we didn't divide by any expression that could be zero; we just used identities and simplified. So all steps are reversible, so all solutions are valid, and we didn't introduce extraneous ones. Therefore, the solution set is indeed x = nπ/5 or x = π/2 + nπ, where n is any integer.But let me check another case. Take x = π/10. Let's compute both sides.Left side: cos(π/10) cos(2*(π/10)) sin(3*(π/10)).Compute each term:cos(π/10) ≈ 0.9511,cos(π/5) ≈ 0.8090,sin(3π/10) ≈ 0.8090.Multiply them: 0.9511 * 0.8090 ≈ 0.7694; then 0.7694 * 0.8090 ≈ 0.6223.Right side: 0.25 sin(2*(π/10)) = 0.25 sin(π/5) ≈ 0.25 * 0.5878 ≈ 0.14695.So left ≈ 0.6223, right ≈ 0.14695. Not equal, so x=π/10 is not a solution. Which is okay, since π/10 = (π/5)/2, but according to our solution, x = nπ/5. For n=1/2, but n must be integer, so x=π/10 is not in the solution set. So that's correct. So the solution set is x = nπ/5 or x = π/2 + nπ. So x=0, π/5, 2π/5, etc., and x=π/2, 3π/2, etc.Wait, but let me check x=2π/5.Left side: cos(2π/5) cos(4π/5) sin(6π/5).Compute each term:cos(2π/5) ≈ 0.3090,cos(4π/5) ≈ -0.8090,sin(6π/5) ≈ sin(π + π/5) = -sin(π/5) ≈ -0.5878.Multiply them: 0.3090 * (-0.8090) ≈ -0.25, then * (-0.5878) ≈ 0.25 * 0.5878 ≈ 0.14695.Right side: 0.25 sin(4π/5) ≈ 0.25 * 0.5878 ≈ 0.14695. So they match. So x=2π/5 is a solution. Similarly, x=3π/5: check if it's a solution.Left side: cos(3π/5) cos(6π/5) sin(9π/5). Wait, cos(3π/5) ≈ -0.8090, cos(6π/5) ≈ -0.3090, sin(9π/5) ≈ sin(π - π/5) = sin(π/5) ≈ 0.5878. Multiply them: (-0.8090)*(-0.3090) ≈ 0.25, then *0.5878 ≈ 0.14695. Right side: 0.25 sin(6π/5) ≈ 0.25*(-0.5878) ≈ -0.14695. Wait, left side ≈0.14695, right side ≈-0.14695. So they are not equal. Therefore, x=3π/5 is not a solution.Wait, that's contradictory. According to our solution, x=3π/5 is n=3, so x=3π/5 should be a solution. But here, left side is positive, right side is negative. So this is a problem. Therefore, x=3π/5 is not a solution. So why is that?Hold on, so perhaps our general solution is missing something. Wait, x=3π/5 is a solution to sin5x=0. Let's check sin5x when x=3π/5. 5x=3π. sin3π=0. So sin5x=0, so according to our solution set, x=3π/5 is a solution. But plugging into original equation, it gives left side ≈0.14695 and right side≈-0.14695. Which are not equal. Therefore, x=3π/5 is not a solution. So our general solution includes some extraneous solutions. That is a problem.Therefore, there must be an error in the solving process. Let me retrace.Original equation: cosx cos2x sin3x = 0.25 sin2x.We transformed the left side to (sin6x + sin4x + sin2x)/4. Then equate to 0.25 sin2x. Then multiply both sides by 4: sin6x + sin4x + sin2x = sin2x. Subtract sin2x: sin6x + sin4x = 0. Then, applied sum-to-product: 2 sin5x cosx =0. Therefore, sin5x=0 or cosx=0.But when x=3π/5, sin5x=sin3π=0. So according to the equation after substitution, sin6x + sin4x =0. Let's check sin6x + sin4x when x=3π/5.Compute sin6x: 6x=18π/5=3π + 3π/5. So sin(18π/5)=sin(3π + 3π/5)=sin(3π/5 + π)= -sin(3π/5)≈-0.5878.sin4x:4x=12π/5=2π + 2π/5. sin(12π/5)=sin(2π/5)≈0.5878.So sin6x + sin4x= -0.5878 + 0.5878=0. So indeed, sin6x + sin4x=0. So equation holds here. But when plugging into original equation, it's not valid. So contradiction. Therefore, our transformation is correct, but plugging into original equation gives inconsistency. Therefore, why?Wait, let's compute original equation at x=3π/5.Left side: cosx cos2x sin3x. x=3π/5.cos(3π/5)=cos(108°)=cos(π - 2π/5)= -cos(2π/5)≈-0.3090.cos2x=cos(6π/5)=cos(π + π/5)= -cos(π/5)≈-0.8090.sin3x=sin(9π/5)=sin(2π - π/5)= -sin(π/5)≈-0.5878.Multiply them: (-0.3090)*(-0.8090)*(-0.5878). Let's compute step by step:First, (-0.3090)*(-0.8090)=0.3090*0.8090≈0.25.Then, 0.25*(-0.5878)= -0.14695.Right side: 0.25 sin2x. 2x=6π/5, sin6π/5≈-0.5878. So 0.25*(-0.5878)= -0.14695.Therefore, left side= -0.14695, right side= -0.14695. So they are equal. Wait, in my previous calculation, I thought sin9π/5≈0.5878, but actually sin9π/5=sin(π + 4π/5)= -sin(4π/5)= -0.5878. Wait, but 9π/5 is 2π - π/5, so sin9π/5= -sin(π/5). Which is approximately -0.5878. So cosx cos2x sin3x= (-0.3090)*(-0.8090)*(-0.5878)= (-0.3090*0.8090*0.5878)= let's compute 0.3090*0.8090≈0.25, then 0.25*0.5878≈0.14695, but with three negatives: cosx is negative, cos2x is negative, sin3x is negative. So three negatives: (-1)^3=-1. So total is -0.14695. Right side is 0.25 sin2x=0.25*sin6π/5=0.25*(-0.5878)= -0.14695. So they are equal. So x=3π/5 is a valid solution.Wait, in my previous check, I made a mistake in the sign of sin3x. So x=3π/5 is indeed a valid solution. So my mistake was in miscalculating sin9π/5. So x=3π/5 is actually a valid solution. Therefore, the solution set is correct.Another example: x=π. Then x=π. Check original equation.Left side: cosπ cos2π sin3π= (-1)(1)(0)=0.Right side:0.25 sin2π=0.25*0=0. So 0=0. Valid. x=π is included in the solution set as x=nπ/5 with n=5.Wait, x=π is n=5: 5π/5=π. So yes. So all these solutions are okay.Another example: x=π/2. As checked before, left side=0, right side=0. So valid.Therefore, conclusion: the general solution is x = nπ/5 or x = π/2 + nπ, for any integer n.But let me check if there's another way to write this solution. For example, sometimes solutions can be combined. Let's see.x = nπ/5 or x = π/2 + nπ. Let's see if these can be expressed together. For example, π/2 + nπ = (2n +1)π/2. So x=(2n+1)π/2. And x= nπ/5. So the solutions are x = nπ/5 and x= (2n+1)π/2. Therefore, the general solution is the union of these two sets.But in some cases, these could overlap? For example, if there exists integer n and m such that nπ/5 = (2m +1)π/2. Then, n/5 = (2m +1)/2 => 2n =5(2m +1). So 2n=10m +5 => n=5m +5/2. But n must be integer, so 5m +5/2 is not integer. Therefore, no overlap. So the solution set is indeed two separate families.Therefore, the final answer is x = nπ/5 or x= π/2 + nπ, where n is any integer. To present the solutions in a boxed format, we can write:x = boxed{frac{npi}{5}} quad text{or} quad x = boxed{frac{pi}{2} + npi}, where n is any integer.But the problem says "put your final answer within boxed{}". However, usually for trigonometric equations with multiple solutions, we present the general solution in terms of n. Since the problem is in Chinese (judging by the original user's input), but the instruction here is in English, but regardless, mathematical notation is universal. The standard way is to present all solutions, so combining both cases.Alternatively, if possible, express both families in a single expression if they can be combined, but here they can't. So the answer is two separate general solutions. Therefore, box both expressions.But in some textbooks, they might write all solutions together as x = frac{npi}{5} or x = frac{pi}{2} + npi for any integer n. So boxed:boxed{x = frac{npi}{5} text{ or } x = frac{pi}{2} + npi text{, where } n in mathbb{Z}}But since the problem might expect specific solutions within a certain interval? Wait, the original problem didn't specify an interval. It just says "solve the equation". So the general solution is as above.Alternatively, if the problem expects the answer in a specific format, but as per instructions, just put the final answer within boxed{}.But the user wrote: "put your final answer within boxed{}". So likely, the answer is the general solution written as two cases. But how to box multiple answers? Maybe separate boxes? But standard practice is to use a single box. So perhaps:boxed{x = frac{npi}{5} text{ or } x = frac{pi}{2} + npi} text{ for any integer } n.But since instructions say to put the final answer within boxed{}, probably including the "for any integer n" inside the box. But LaTeX-wise, it's better to write:boxed{x = frac{npi}{5} text{ or } x = frac{pi}{2} + npi, n in mathbb{Z}}.Yes, that's concise and standard.Final AnswerThe solutions are boxed{x = frac{npi}{5} text{ or } x = frac{pi}{2} + npi, n in mathbb{Z}}.</think>To solve the equation (cos x cos 2x sin 3x = 0.25 sin 2x), we start by using trigonometric identities to simplify both sides. First, we apply the product-to-sum identity to (cos x cos 2x):[cos x cos 2x = frac{cos(3x) + cos x}{2}]Substituting this back into the left-hand side of the original equation, we get:[frac{cos(3x) + cos x}{2} sin 3x]Distributing (sin 3x) gives:[frac{cos 3x sin 3x + cos x sin 3x}{2}]Using the product-to-sum identities again on each term:[cos 3x sin 3x = frac{sin 6x}{2} quad text{and} quad cos x sin 3x = frac{sin 4x + sin 2x}{2}]Thus, the left-hand side becomes:[frac{frac{sin 6x}{2} + frac{sin 4x + sin 2x}{2}}{2} = frac{sin 6x + sin 4x + sin 2x}{4}]The original equation then simplifies to:[frac{sin 6x + sin 4x + sin 2x}{4} = frac{sin 2x}{4}]Multiplying both sides by 4 and subtracting (sin 2x):[sin 6x + sin 4x = 0]Using the sum-to-product identity:[sin 6x + sin 4x = 2 sin 5x cos x = 0]This gives two cases:1. (sin 5x = 0) which implies (5x = npi) or (x = frac{npi}{5})2. (cos x = 0) which implies (x = frac{pi}{2} + npi)Both cases are checked for validity and no extraneous solutions are found. Therefore, the general solution is:[boxed{x = frac{npi}{5} text{ or } x = frac{pi}{2} + npi, n in mathbb{Z}}],<think>Okay, so I need to solve the equation cos x cos 2x sin 3x = 0.25 sin 2x. Hmm, trigonometric equation. Let me see how to approach this. First, maybe I can rearrange the equation to bring all terms to one side. Let's try that.Starting with the original equation:cos x cos 2x sin 3x - 0.25 sin 2x = 0Hmm, not sure if that helps immediately. Maybe I can factor something out? Let's look for common factors. The left side has terms with cos x, cos 2x, sin 3x, and the right side term is sin 2x. Maybe I can express sin 3x in terms of sin 2x and other terms? Or perhaps use some product-to-sum formulas here?Wait, product-to-sum identities might be useful here because we have products of trigonometric functions. Let me recall the product-to-sum identities. For example, cos A cos B can be written as [cos(A+B) + cos(A-B)] / 2. Similarly, sin A cos B is [sin(A+B) + sin(A-B)] / 2. Maybe applying these identities step by step could simplify the left-hand side.Let me first look at cos x cos 2x. Let's apply the product-to-sum formula here:cos x cos 2x = [cos(x + 2x) + cos(x - 2x)] / 2 = [cos 3x + cos(-x)] / 2. Since cos(-x) = cos x, this simplifies to [cos 3x + cos x] / 2.So substituting back into the original equation:[cos 3x + cos x]/2 * sin 3x = 0.25 sin 2xMultiply both sides by 2 to eliminate the denominator:(cos 3x + cos x) sin 3x = 0.5 sin 2xHmm, so now we have (cos 3x + cos x) sin 3x. Maybe I can distribute sin 3x here:cos 3x sin 3x + cos x sin 3x = 0.5 sin 2xAgain, looking at terms like cos 3x sin 3x and cos x sin 3x. Let's apply product-to-sum identities again.First term: cos 3x sin 3x. Using the identity sin A cos A = (1/2) sin 2A. So cos 3x sin 3x = (1/2) sin 6x.Second term: cos x sin 3x. Using identity sin A cos B = [sin(A+B) + sin(A-B)] / 2. So cos x sin 3x = [sin(3x + x) + sin(3x - x)] / 2 = [sin 4x + sin 2x] / 2.Putting it all back together:(1/2) sin 6x + [sin 4x + sin 2x]/2 = 0.5 sin 2xMultiply both sides by 2 to eliminate denominators:sin 6x + sin 4x + sin 2x = sin 2xSubtract sin 2x from both sides:sin 6x + sin 4x = 0Hmm, so now we have sin 6x + sin 4x = 0. Maybe use sum-to-product here. The identity sin A + sin B = 2 sin[(A+B)/2] cos[(A-B)/2].Applying that:sin 6x + sin 4x = 2 sin[(6x + 4x)/2] cos[(6x - 4x)/2] = 2 sin 5x cos xSo 2 sin 5x cos x = 0Therefore, the equation simplifies to 2 sin 5x cos x = 0. So either sin 5x = 0 or cos x = 0.Let's solve each case separately.Case 1: cos x = 0Solutions are x = π/2 + πk, where k is any integer.Case 2: sin 5x = 0Solutions are 5x = πn, so x = πn/5, where n is any integer.But wait, we need to check if all these solutions are valid, because when we performed algebraic manipulations, sometimes extraneous solutions can be introduced. Let's verify.Original equation: cos x cos 2x sin 3x = 0.25 sin 2xFirst, check solutions from cos x = 0, which are x = π/2 + πk.Plug x = π/2 into the equation:cos(π/2) = 0, so left-hand side is 0. Right-hand side: 0.25 sin(2*(π/2)) = 0.25 sin π = 0. So 0 = 0, which holds.Similarly for x = 3π/2, cos(3π/2) = 0, left-hand side 0. Right-hand side: 0.25 sin(3π) = 0. So all these solutions are valid.Now check solutions from sin 5x = 0, which are x = πn/5.Plug x = πn/5 into the original equation.Let me take x = 0 (n=0). Then cos 0 = 1, cos 0 = 1, sin 0 = 0. Left-hand side: 1*1*0 = 0. Right-hand side: 0.25 sin 0 = 0. So 0=0, valid.x = π/5 (n=1):cos(π/5) ≈ 0.8090, cos(2π/5) ≈ 0.3090, sin(3π/5) ≈ 0.5878. Multiply these: 0.8090 * 0.3090 * 0.5878 ≈ 0.147. Right-hand side: 0.25 sin(2π/5) ≈ 0.25 * 0.5878 ≈ 0.1469. So approximately equal, which holds. So this solution is valid.x = 2π/5 (n=2):cos(2π/5) ≈ 0.3090, cos(4π/5) ≈ -0.8090, sin(6π/5) ≈ -0.5878. Multiply left-hand side: 0.3090 * (-0.8090) * (-0.5878) ≈ 0.3090 * 0.8090 * 0.5878 ≈ 0.147. Right-hand side: 0.25 sin(4π/5) ≈ 0.25 * 0.5878 ≈ 0.1469. Again, approximately equal. Valid.x = π (n=5):cos π = -1, cos 2π = 1, sin 3π = 0. Left-hand side: (-1)(1)(0) = 0. Right-hand side: 0.25 sin 2π = 0. Valid.x = π/2 (n=5/2, but n must be integer). Wait, x = πn/5, so n must be integer. So x = π/2 would correspond to n=5/2, which is not integer, hence x=π/2 is covered in the first case (cos x =0). So overlapping solutions are at x=π/2 + πk, which are included in both cases? Wait, if x = π/2, then in sin5x: sin(5*(π/2)) = sin(5π/2) = 1 ≠0. So x=π/2 is not a solution from sin5x=0. Therefore, the two cases are distinct.Wait, but when n=5k in sin5x=0, then x=πk. But in this case, x=πk is also part of the general solution. But x=πk when k is even or odd. For example, x=0, π, 2π, etc. Let's check x=π. From sin5x=0, x=π is a solution. From cosx=0, x=π/2 + πk, which at k=0.5 would be x=π, but k must be integer. So x=π is not in the first case. Wait, no. cos x=0 solutions are x=π/2 + πk. So x=π/2, 3π/2, 5π/2, etc. So x=π is not in the first case, but x=π is a solution from sin5x=0 (n=5). Therefore, the solutions from sin5x=0 include x=0, π/5, 2π/5, 3π/5, 4π/5, π, 6π/5, etc. So some of these may or may not coincide with cosx=0.But since we already considered the two cases separately, and x=π/2 + πk (from cosx=0) do not overlap with x=πn/5 except when π/2 + πk = πn/5. Let's see if that's possible. For example, π/2 = πn/5 ⇒ n=5/2, which is not integer. So there is no overlap. Therefore, all solutions are either x=π/2 + πk or x=πn/5. Wait, but in x=πn/5, when n is a multiple of 5, like n=5, x=π. But cos π = -1 ≠0, so not in the first case. So all solutions are separate.Therefore, combining both cases, the general solution is x=π/2 + πk or x=πn/5 for integers k and n.But wait, we need to check if all solutions from sin5x=0 satisfy the original equation, or if some of them are extraneous. Let me check x=π/5. As before, when plugging x=π/5, the left-hand side was approximately 0.147 and the right-hand side was approximately 0.1469, which is due to rounding, but mathematically exact?Let me compute more precisely. Let's compute cos(π/5) and cos(2π/5) and sin(3π/5):cos(π/5) = (sqrt(5)+1)/4 * 2 = (sqrt(5)+1)/4 * 2? Wait, actually, cos(π/5) is equal to (1 + sqrt(5))/4 * 2. Let me recall exact value:cos(π/5) = (1 + sqrt(5))/4 * 2 = [sqrt(5) + 1]/4 * 2? Wait, no. Let me recall that cos(36°) = (1 + sqrt(5))/4 * 2. Wait, better to recall exact value:cos(π/5) = (1 + sqrt(5))/4 * 2? Wait, actually, it's known that cos(π/5) = (1 + sqrt(5))/4 * 2 is incorrect. Let me check exact value.Actually, cos(36°) = cos(π/5) = (sqrt(5) + 1)/4 * 2. Wait, let's compute it correctly.We know that cos(36°) = (1 + sqrt(5))/4 * 2, but maybe more accurately:In a regular pentagon, cos(72°) = (sqrt(5)-1)/4 * 2. Wait, perhaps I should use exact identities.Alternatively, we can compute cos(π/5):Let θ = π/5, so 5θ = π. Then 3θ = π - 2θ. So cos(3θ) = cos(π - 2θ) = -cos(2θ). Using the identity cos(3θ) = 4cos^3θ - 3cosθ and cos(2θ) = 2cos^2θ -1. Therefore:4cos^3θ - 3cosθ = - (2cos^2θ -1)Bring all terms to left side:4cos^3θ - 3cosθ + 2cos^2θ -1 =0Let x = cosθ:4x^3 + 2x^2 -3x -1 =0Looking for rational roots, possible roots are ±1, ±1/2, ±1/4. Testing x=1: 4 + 2 -3 -1=2≠0. x=-1: -4 + 2 +3 -1=0. So x=-1 is a root. Then factor out (x +1):Using polynomial division or synthetic division:Divide 4x^3 + 2x^2 -3x -1 by (x +1):Coefficients: 4 | 2 | -3 | -1Using synthetic division for root x=-1:Bring down 4. Multiply by -1: -4. Add to next term: 2 + (-4)= -2. Multiply by -1: 2. Add to next term: -3 +2= -1. Multiply by -1:1. Add to last term: -1 +1=0. So quotient is 4x^2 -2x -1.Therefore, 4x^3 +2x^2 -3x -1= (x +1)(4x^2 -2x -1). Set to zero:(x +1)(4x^2 -2x -1)=0. Solutions are x=-1, and roots of 4x^2 -2x -1=0. Using quadratic formula:x = [2 ± sqrt(4 +16)] /8 = [2 ± sqrt(20)] /8 = [2 ± 2sqrt(5)] /8 = [1 ± sqrt(5)]/4.Since θ=π/5 is acute, cosθ is positive. Therefore, cos(π/5)= [1 + sqrt(5)]/4 *2? Wait, [1 + sqrt(5)]/4 ≈ (1 + 2.236)/4 ≈ 0.809, which matches the approximate value. So exact value is (sqrt(5)+1)/4 *2? Wait, no, [1 + sqrt(5)]/4 is ≈0.809, but actually cos(π/5)= (1 + sqrt(5))/4 *2? Wait, no, wait. Let me check:Wait, the quadratic solution is x = [1 ± sqrt(5)]/4. The positive root is [1 + sqrt(5)]/4 ≈ (1 + 2.236)/4 ≈ 0.809, which is correct. So cos(π/5)= [1 + sqrt(5)]/4 *2? Wait, no, cos(π/5)= [1 + sqrt(5)]/4 multiplied by 2? Wait, no. Wait, the solution is x= [1 + sqrt(5)]/4, so cos(π/5)= [1 + sqrt(5)]/4 * 2? Wait, no, why? The quadratic equation gives x = [1 + sqrt(5)]/4. So cos(π/5)= [1 + sqrt(5)]/4 ≈0.809, which is correct. Similarly, cos(2π/5)= [sqrt(5)-1]/4 ≈0.309.Therefore, cos(π/5) * cos(2π/5) * sin(3π/5) = [ (1 + sqrt(5))/4 ] * [ (sqrt(5) -1)/4 ] * sin(3π/5). Now, sin(3π/5)= sin(2π/5)= sqrt[(5 - sqrt(5))/8]*2≈0.5878. Let me compute the exact product:First, compute [ (1 + sqrt(5))(sqrt(5) -1) ] / (16) = [ (sqrt(5))^2 -1^2 ] /16 = (5 -1)/16 = 4/16 = 1/4. So cos(π/5) cos(2π/5) = 1/4. Then multiply by sin(3π/5)= sin(2π/5)= sqrt[(5 - sqrt(5))/8]*2. Wait, exact value of sin(2π/5)=sqrt( (5 - sqrt(5))/8 )*2. Let me compute that:sin(2π/5)= sqrt( (5 - sqrt(5))/8 )*2. Squaring this: (5 - sqrt(5))/8 *4= (5 - sqrt(5))/2 ≈ (5 - 2.236)/2≈1.382/2≈0.691, which is sin^2(2π/5). So sin(2π/5)≈0.5878, which matches.So cos(π/5) cos(2π/5) sin(3π/5)= (1/4) * sin(2π/5). Then the left-hand side is (1/4) sin(2π/5). The right-hand side is 0.25 sin(2x)=0.25 sin(2*(π/5))=0.25 sin(2π/5). Therefore, both sides are equal. Hence, x=π/5 is indeed a solution.Similarly, x=2π/5: Let's check.cos(2π/5)= [sqrt(5)-1]/4≈0.309, cos(4π/5)=cos(π - π/5)= -cos(π/5)= -[1 + sqrt(5)]/4≈-0.809, sin(6π/5)=sin(π + π/5)= -sin(π/5)= -sqrt[(5 - sqrt(5))/8]*2≈-0.5878. Then left-hand side: 0.309 * (-0.809) * (-0.5878)≈0.309 *0.809 *0.5878≈0.147. Right-hand side:0.25 sin(4π/5)=0.25 sin(π/5)=0.25 *0.5878≈0.1469. So holds true.Similarly, x=3π/5: cos(3π/5)=cos(π - 2π/5)= -cos(2π/5)= -[sqrt(5)-1]/4≈-0.309, cos(6π/5)= wait, cos(2*(3π/5))=cos(6π/5)=cos(π + π/5)= -cos(π/5)≈-0.809, sin(9π/5)=sin(2π - π/5)= -sin(π/5)≈-0.5878. So left-hand side: -0.309 * -0.809 * -0.5878≈-0.309*0.809*0.5878≈-0.147. Right-hand side:0.25 sin(6π/5)=0.25*(-sin(π/5))≈-0.1469. So both sides ≈-0.147≈-0.1469. Valid.x=4π/5: cos(4π/5)= -cos(π/5)≈-0.809, cos(8π/5)=cos(2π - 2π/5)=cos(2π/5)≈0.309, sin(12π/5)=sin(2π + 2π/5)=sin(2π/5)≈0.5878. Left-hand side: -0.809 *0.309 *0.5878≈-0.147. Right-hand side:0.25 sin(8π/5)=0.25 sin(2π - 2π/5)=0.25*(-sin(2π/5))≈-0.1469. Again, valid.x=π: cos π = -1, cos 2π=1, sin 3π=0. Left-hand side: (-1)(1)(0)=0. Right-hand side:0.25 sin 2π=0. Valid.x=6π/5: similar to x=π/5 but shifted by π. Let's see: cos(6π/5)=cos(π + π/5)= -cos(π/5)≈-0.809, cos(12π/5)=cos(2π + 2π/5)=cos(2π/5)≈0.309, sin(18π/5)=sin(3π + 3π/5)=sin(3π/5 + π)= -sin(3π/5)= -sin(2π/5)≈-0.5878. Left-hand side: -0.809 *0.309 * -0.5878≈0.147. Right-hand side:0.25 sin(12π/5)=0.25 sin(2π/5)≈0.1469. Valid.So it seems all solutions from sin5x=0 are valid. Similarly, solutions from cosx=0: x=π/2 +kπ.Check x=π/2:cos(π/2)=0, so left-hand side is 0. Right-hand side:0.25 sin(π)=0. Valid.x=3π/2: same thing, both sides 0. Valid.So all solutions are indeed valid.Therefore, the general solution is x=π/2 +kπ or x=πn/5 for integers k and n. But wait, x=πn/5 includes x=π/2 when n=5/2, but n must be integer, so x=π/2 is not included in x=πn/5. Therefore, the solutions are two separate sets:1. x = π/2 + kπ, k ∈ ℤ2. x = πn/5, n ∈ ℤBut we need to check if these two sets have any overlap. Suppose there exists integers k and n such that π/2 + kπ = πn/5. Then dividing both sides by π: 1/2 +k = n/5 ⇒ n =5(k +1/2). But n must be integer, so 5(k +1/2) must be integer. However, 5(k +1/2)=5k +2.5, which is not integer. Therefore, there's no overlap between the two solution sets. So all solutions are distinct.Therefore, the complete solution is:x = π/2 + kπ or x = πn/5, where k, n are integers.But wait, let's check x=π. It is included in the second case when n=5: x=π*5/5=π. So x=π is a solution from the second case. Similarly, x=0 is included in the second case when n=0. So x=0 is a solution.So the solutions can be written as:x = π/2 + kπ or x = πn/5 for any integers k and n.But since n is any integer, the second case covers all multiples of π/5, which includes x=0, π/5, 2π/5, etc., both positive and negative.Hence, combining both cases, the solutions are:All real numbers x such that x = π/2 + kπ or x = πn/5 where k and n are integers.Therefore, the final answer should be presented as the union of these two sets. In boxed notation, we can write:x = boxed{dfrac{pi}{2} + kpi} quad text{and} quad x = boxed{dfrac{pi n}{5}} quad text{for any integer } k, n.But maybe the problem expects the solutions combined in a particular way. Alternatively, sometimes it's written as:x = dfrac{pi}{2} + kpi quad text{or} quad x = dfrac{npi}{5}, quad k, n in mathbb{Z}So in boxed form, since the question says "put your final answer within boxed{}", perhaps list both solutions in one box, separated by commas or something. But typically, multiple solutions are boxed separately. Wait, looking at the problem statement: "put your final answer within boxed{}". So likely, they expect the general solution expressed as two separate cases in one box. For example:boxed{x = dfrac{pi}{2} + kpi text{ or } x = dfrac{npi}{5}, text{ where } k, n in mathbb{Z}}But I need to check if the original equation had any restrictions, but since none were mentioned, and all solutions are valid, this should be the answer.Final AnswerThe solutions are boxed{x = dfrac{pi}{2} + kpi} text{ and } boxed{x = dfrac{npi}{5}} for any integers ( k ) and ( n ).Wait, the user instruction says "put your final answer within boxed{}". So maybe need to combine both solutions into a single box. But standard practice is to box all solutions. If multiple solutions, box each or box together. Since in Chinese practice, sometimes they expect multiple boxed answers, but here the system might accept multiple boxes. However, looking at the problem number 8.211., this might be from a textbook where the answer is expected in two parts. But as the user instruction says "put your final answer within boxed{}", perhaps they want all solutions combined in one box. Let me check with another problem.Alternatively, maybe the problem expects specific solutions within a certain interval, but since none is specified, it's general solutions.Therefore, combining both cases into one box:boxed{x = dfrac{pi}{2} + kpi text{ or } x = dfrac{npi}{5}, text{ where } k, n in mathbb{Z}}Yes, that seems appropriate. So the final answer is as above.Final Answerboxed{x = dfrac{pi}{2} + kpi} text{ and } boxed{x = dfrac{npi}{5}} text{ for integers } k text{ and } n.But again, the instruction says "put your final answer within boxed{}". Maybe two separate boxed answers. However, the system might not accept multiple boxes. Hmmm. Wait, the original problem is numbered 8.211, which might imply that it's a single answer. Looking back, when they say "put your final answer within boxed{}", maybe it's expecting all solutions together in one box. So perhaps:boxed{x = dfrac{pi}{2} + kpi text{ or } x = dfrac{npi}{5}, text{ where } k, n text{ are integers}}Yes, this is a single box containing all solutions.Final Answerboxed{x = dfrac{pi}{2} + kpi text{ or } x = dfrac{npi}{5}, text{ where } k, n in mathbb{Z}}</think>To solve the equation (cos x cos 2x sin 3x = 0.25 sin 2x), we start by rearranging and applying trigonometric identities.1. Rearrange the equation: [ cos x cos 2x sin 3x - 0.25 sin 2x = 0 ]2. Apply product-to-sum identities: - For (cos x cos 2x): [ cos x cos 2x = frac{cos 3x + cos x}{2} ] - Substitute back into the equation: [ left( frac{cos 3x + cos x}{2} right) sin 3x = 0.25 sin 2x ] - Multiply both sides by 2: [ (cos 3x + cos x) sin 3x = 0.5 sin 2x ]3. Distribute (sin 3x): - Use product-to-sum identities again: - For (cos 3x sin 3x): [ cos 3x sin 3x = frac{1}{2} sin 6x ] - For (cos x sin 3x): [ cos x sin 3x = frac{sin 4x + sin 2x}{2} ] - Combine the terms: [ frac{1}{2} sin 6x + frac{sin 4x + sin 2x}{2} = 0.5 sin 2x ] - Multiply by 2 and simplify: [ sin 6x + sin 4x + sin 2x = sin 2x ] [ sin 6x + sin 4x = 0 ]4. Apply sum-to-product identity: - Use (sin A + sin B = 2 sin frac{A+B}{2} cos frac{A-B}{2}): [ 2 sin 5x cos x = 0 ]5. Solve the equation: - Either (sin 5x = 0) or (cos x = 0): - (cos x = 0) gives solutions (x = frac{pi}{2} + kpi) for integers (k). - (sin 5x = 0) gives solutions (x = frac{npi}{5}) for integers (n).6. Verify solutions: - All solutions from both cases are valid and do not overlap.Thus, the general solutions are:[boxed{x = dfrac{pi}{2} + kpi text{ or } x = dfrac{npi}{5}, text{ where } k, n in mathbb{Z}}]]

🔑:## Solution.Let's write the equation as:cos x cos 2 x sin 3 x - 0.5 sin x cos x = 0 Leftrightarrow cos x(cos 2 x sin 3 x - 0.5 sin x) = 0.## From this, eithercos x = 0, x_{1} = frac{pi}{2} + pi k = frac{pi}{2}(2 k + 1), k in mathbb{Z},## orcos 2 x sin 3 x - 0.5 sin x = 0, or 2 cos 2 x sin 3 x - sin x = 0.Using the formula sin alpha cos beta = frac{1}{2}(sin (alpha - beta) + sin (alpha + beta)), we have:sin (3 x - 2 x) + sin (3 x + 2 x) - sin x = 0 Leftrightarrow sin x + sin 5 x - sin x = 0 LeftrightarrowLeftrightarrow sin 5 x, 5 x = pi n, x_{2} = frac{pi n}{5}, n in mathbb{Z}.Answer: x_{1} = frac{pi}{2}(2 k + 1), x_{2} = frac{pi n}{5}, where k and n in mathbb{Z}.

🧠:[<think>Okay, so I need to find the maximum and minimum values of the function f(x) = (2x² + √2 sin(x + π/4)) / (2x² + cos x), and then add them together to get a + b. Hmm, let's start by understanding the function.First, both the numerator and the denominator are combinations of quadratic terms and trigonometric functions. Since the quadratic terms are 2x² in both the numerator and denominator, maybe they dominate as x becomes large. Let me check the behavior as x approaches infinity.As x → ∞, the 2x² terms will dominate over the sin and cos terms because sin and cos oscillate between -1 and 1. So, both the numerator and denominator will be approximately 2x², so the function f(x) should approach 1. But near x = 0, the trigonometric terms might have a bigger impact. Let me check f(0).At x = 0:Numerator: 2(0)² + √2 sin(0 + π/4) = √2 sin(π/4) = √2*(√2/2) = 1Denominator: 2(0)² + cos(0) = 1So f(0) = 1/1 = 1. Interesting, so at x=0, it's also 1.What about x = π/4? Let me try that.At x = π/4:Numerator: 2*(π/4)² + √2 sin(π/4 + π/4) = 2*(π²/16) + √2 sin(π/2) = π²/8 + √2*1 ≈ (9.87)/8 + 1.414 ≈ 1.234 + 1.414 ≈ 2.648Denominator: 2*(π/4)^2 + cos(π/4) = 2*(π²/16) + √2/2 ≈ 1.234 + 0.707 ≈ 1.941So f(π/4) ≈ 2.648 / 1.941 ≈ 1.364. So that's higher than 1. So maybe the maximum is around there?Wait, but maybe it's better to take the derivative and find critical points. Since the function is a quotient, I can use the quotient rule for differentiation.Let me denote numerator as N(x) = 2x² + √2 sin(x + π/4)Denominator as D(x) = 2x² + cos xThen f'(x) = [N'(x) D(x) - N(x) D'(x)] / [D(x)]²First, compute N'(x):N'(x) = d/dx [2x² + √2 sin(x + π/4)] = 4x + √2 cos(x + π/4) * 1 = 4x + √2 cos(x + π/4)D'(x) = d/dx [2x² + cos x] = 4x - sin xTherefore, f'(x) = [ (4x + √2 cos(x + π/4))(2x² + cos x) - (2x² + √2 sin(x + π/4))(4x - sin x) ] / (2x² + cos x)^2This looks complicated. Maybe simplify the numerator?Let me expand the numerator:First term: (4x + √2 cos(x + π/4))(2x² + cos x)= 4x*2x² + 4x*cos x + √2 cos(x + π/4)*2x² + √2 cos(x + π/4)*cos x= 8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos xSecond term: -(2x² + √2 sin(x + π/4))(4x - sin x)= - [2x²*4x - 2x² sin x + √2 sin(x + π/4)*4x - √2 sin(x + π/4) sin x]= - [8x³ - 2x² sin x + 4√2 x sin(x + π/4) - √2 sin(x + π/4) sin x]So total numerator:[8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos x] - [8x³ - 2x² sin x + 4√2 x sin(x + π/4) - √2 sin(x + π/4) sin x]Let's distribute the negative sign:8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos x -8x³ + 2x² sin x -4√2 x sin(x + π/4) + √2 sin(x + π/4) sin xNow, combine like terms:8x³ -8x³ = 04x cos x remains.2√2 x² cos(x + π/4) + 2x² sin x√2 cos(x + π/4) cos x remains-4√2 x sin(x + π/4) remains+√2 sin(x + π/4) sin x remainsSo, the numerator simplifies to:4x cos x + 2√2 x² cos(x + π/4) + 2x² sin x + √2 cos(x + π/4) cos x -4√2 x sin(x + π/4) + √2 sin(x + π/4) sin xHmm, this is still quite complex. Maybe we can factor some terms or use trigonometric identities.Note that x + π/4 is the argument inside the sin and cos in the numerator. Let me recall that sin(x + π/4) = sin x cos π/4 + cos x sin π/4 = (sin x + cos x)/√2. Similarly, cos(x + π/4) = cos x cos π/4 - sin x sin π/4 = (cos x - sin x)/√2.Let me substitute these into the expression to see if that helps.First, cos(x + π/4) = (cos x - sin x)/√2Similarly, sin(x + π/4) = (sin x + cos x)/√2So replace all instances of cos(x + π/4) and sin(x + π/4) with these expressions.Starting with the numerator:4x cos x + 2√2 x² [ (cos x - sin x)/√2 ] + 2x² sin x + √2 [ (cos x - sin x)/√2 ] cos x -4√2 x [ (sin x + cos x)/√2 ] + √2 [ (sin x + cos x)/√2 ] sin xSimplify each term step by step.First term: 4x cos x remains.Second term: 2√2 x² * (cos x - sin x)/√2 = 2√2 / √2 x² (cos x - sin x) = 2x² (cos x - sin x)Third term: 2x² sin x remains.Fourth term: √2 * (cos x - sin x)/√2 * cos x = (cos x - sin x) cos xFifth term: -4√2 x * (sin x + cos x)/√2 = -4√2 / √2 x (sin x + cos x) = -4x (sin x + cos x)Sixth term: √2 * (sin x + cos x)/√2 * sin x = (sin x + cos x) sin xNow, let's combine all terms:4x cos x + 2x² (cos x - sin x) + 2x² sin x + (cos x - sin x) cos x -4x (sin x + cos x) + (sin x + cos x) sin xLet's simplify term by term.First, 4x cos x.Second term: 2x² cos x - 2x² sin xThird term: +2x² sin x, so combining with the second term: 2x² cos x -2x² sin x +2x² sin x = 2x² cos xFourth term: (cos x - sin x) cos x = cos² x - sin x cos xFifth term: -4x sin x -4x cos xSixth term: sin² x + sin x cos xNow, combine all these:First term: 4x cos xSecond and third term combined: 2x² cos xFourth term: cos² x - sin x cos xFifth term: -4x sin x -4x cos xSixth term: sin² x + sin x cos xNow, let's combine like terms:4x cos x -4x cos x = 0 (from first and fifth term)Similarly, 4x cos x -4x cos x cancels.Looking at the x terms: -4x sin x from fifth term.Wait, let me list each part again:Starting with:4x cos x (first term)2x² cos x (from second and third terms)cos² x - sin x cos x (fourth term)-4x sin x -4x cos x (fifth term)sin² x + sin x cos x (sixth term)So:4x cos x -4x cos x = 0Then, -4x sin x remains.2x² cos x remains.cos² x - sin x cos x + sin² x + sin x cos xThe - sin x cos x and + sin x cos x cancel.So cos² x + sin² x = 1Thus, the numerator simplifies to:2x² cos x -4x sin x + 1Wait, let me check again:From fourth and sixth terms:cos² x - sin x cos x + sin² x + sin x cos x = cos² x + sin² x + (- sin x cos x + sin x cos x) = 1 + 0 = 1So the numerator becomes:(2x² cos x) + (-4x sin x) + 1Thus, f'(x) = [2x² cos x -4x sin x + 1] / (2x² + cos x)^2So critical points occur when the numerator is zero:2x² cos x -4x sin x + 1 = 0Hmm, this equation seems complicated. Let me see if I can solve it or find possible solutions.Alternatively, maybe there's a symmetry or substitution. Let's think.First, note that f(x) has even symmetry? Let's check f(-x):Numerator: 2(-x)^2 + √2 sin(-x + π/4) = 2x² + √2 sin(-x + π/4) = 2x² + √2 sin(π/4 - x)Similarly, sin(π/4 - x) = sin π/4 cos x - cos π/4 sin x = (√2/2)(cos x - sin x)So numerator becomes 2x² + √2*(√2/2)(cos x - sin x) = 2x² + (cos x - sin x)Denominator: 2(-x)^2 + cos(-x) = 2x² + cos xSo f(-x) = [2x² + cos x - sin x] / (2x² + cos x) = 1 - [sin x / (2x² + cos x)]But the original function f(x) is [2x² + √2 sin(x + π/4)] / (2x² + cos x). Let's see if they are related.Wait, perhaps if we let x = t - π/4, but not sure.Alternatively, maybe we can write the numerator in terms of sin(x + π/4):We know that sin(x + π/4) = (sin x + cos x)/√2, so √2 sin(x + π/4) = sin x + cos x. Therefore, the numerator is 2x² + sin x + cos x. So f(x) = (2x² + sin x + cos x) / (2x² + cos x) = 1 + [sin x / (2x² + cos x)]Ah, that's a good simplification! So f(x) = 1 + [sin x / (2x² + cos x)]Therefore, f(x) = 1 + sinx / (2x² + cosx). That might be easier to work with.So then, f(x) = 1 + sinx / (2x² + cosx). So to find the maximum and minimum, we can look at the term sinx / (2x² + cosx). Let me denote g(x) = sinx / (2x² + cosx). Then f(x) = 1 + g(x). Therefore, maximum of f(x) is 1 + max(g(x)), and minimum is 1 + min(g(x)). So a + b = (1 + max(g)) + (1 + min(g)) = 2 + max(g) + min(g). So if I can find max(g) + min(g), then add 2. But maybe that's not helpful. Alternatively, perhaps max(g) + min(g) = 0? Let's see.Wait, g(x) is an odd function? Let's check. Let's check g(-x):g(-x) = sin(-x) / (2(-x)^2 + cos(-x)) = -sinx / (2x² + cosx) = -g(x). So yes, g(x) is odd. Therefore, if x is a critical point where g(x) is maximum, then -x is a critical point where g(-x) = -g(x), which would be the minimum. Therefore, max(g) + min(g) = max(g) - max(g) = 0. Therefore, a + b = 2 + 0 = 2. Wait, is that possible?Wait, but is the function g(x) odd? Let's confirm again.g(-x) = sin(-x)/(2x² + cos(-x)) = (-sinx)/(2x² + cosx) = -g(x). Therefore, yes, g(x) is odd. Therefore, it's symmetric about the origin. So if the maximum value is M, then the minimum is -M. Therefore, a + b = (1 + M) + (1 - M) = 2. Therefore, regardless of what M is, a + b = 2. So maybe the answer is 2?But wait, the problem states that "the maximum and minimum values are a, b respectively, then a + b =". So if the maximum value is a and the minimum is b, then a + b = 2? That seems too straightforward. But perhaps that's the case.But let me check with the example I did earlier. At x = π/4, f(x) ≈ 1.364, which is higher than 1, and if the function is 1 + g(x), and g(x) is odd, then if there's a maximum at some x, there's a minimum at -x. But wait, in my previous example, x = π/4 gave a higher value. If x = -π/4, what would f(x) be?At x = -π/4:Numerator: 2*(π²/16) + √2 sin(-π/4 + π/4) = same as before, 2*(π²/16) + √2 sin(0) = π²/8 ≈ 1.234Denominator: 2*(π²/16) + cos(-π/4) = π²/8 + √2/2 ≈ 1.234 + 0.707 ≈ 1.941So f(-π/4) ≈ 1.234 / 1.941 ≈ 0.636. Then, since f(x) = 1 + sinx/(2x² + cosx), so sinx at x = -π/4 is -√2/2, so sinx/(denominator) would be negative, so f(-π/4) = 1 - something. So 0.636 ≈ 1 - 0.364, which seems reasonable.But if the maximum is 1.364 and the minimum is 0.636, then a + b ≈ 2, which is exactly 1.364 + 0.636 ≈ 2. So that seems to check out.But let's confirm if the maximum and minimum are indeed symmetric around 1. If the function is f(x) = 1 + g(x), and g(x) is odd, then yes, the maximum and minimum of f(x) would be symmetric around 1. Therefore, a + b = 2. So maybe the answer is 2?But wait, I need to be careful here. The function g(x) is odd, which means that g(-x) = -g(x). So if x = c is where g(x) reaches its maximum, then x = -c is where g(x) reaches its minimum. Therefore, the maximum of f(x) is 1 + g(c), and the minimum is 1 + g(-c) = 1 - g(c). Therefore, a + b = (1 + g(c)) + (1 - g(c)) = 2. Therefore, regardless of the value of g(c), the sum a + b is always 2.But is this always the case? Let me check if the function f(x) indeed has this property.Suppose that f(x) can be written as 1 + g(x), where g(x) is an odd function. Then, yes, the maximum and minimum of f(x) would be 1 + M and 1 - M, so their sum is 2. But does this hold even if the maximum and minimum are not achieved at symmetric points? For example, could the maximum be achieved at some point c, but the minimum not at -c?However, since g(x) is odd, then if c is a critical point where g(c) is maximum, then -c is a critical point where g(-c) = -g(c) is minimum. Because the derivative of an odd function is even, so critical points would come in pairs c and -c. Therefore, the maximum and minimum would be symmetric. Therefore, the conclusion holds.Therefore, the answer is 2. Therefore, a + b = 2.But let me double-check with another example. Take x = π/2.At x = π/2:Numerator: 2*(π/2)^2 + √2 sin(π/2 + π/4) = 2*(π²/4) + √2 sin(3π/4) = π²/2 + √2*(√2/2) = π²/2 + 1 ≈ 4.934 + 1 ≈ 5.934Denominator: 2*(π/2)^2 + cos(π/2) = π²/2 + 0 ≈ 4.934Therefore, f(π/2) ≈ 5.934 / 4.934 ≈ 1.203Which is less than the value at π/4. So maximum might indeed be around π/4.Similarly, at x = -π/2:Numerator: 2*(π/2)^2 + √2 sin(-π/2 + π/4) = π²/2 + √2 sin(-π/4) = π²/2 - √2*(√2/2) = π²/2 - 1 ≈ 4.934 - 1 ≈ 3.934Denominator: 2*(π/2)^2 + cos(-π/2) = π²/2 + 0 ≈ 4.934Thus, f(-π/2) ≈ 3.934 / 4.934 ≈ 0.797Again, f(-π/2) ≈ 0.797, which is 1 - 0.203, so indeed, if the maximum at some point is 1.364, the minimum would be around 0.636, adding to 2.Alternatively, since f(x) = 1 + sinx/(2x² + cosx), then the maximum value occurs when sinx/(2x² + cosx) is maximized, and the minimum when it's minimized (i.e., most negative). Since the function sinx/(2x² + cosx) is odd, its maximum and minimum are negatives of each other. Therefore, their sum cancels out, leaving a + b = 2.Therefore, the answer should be 2.Final Answerboxed{2}</think>Given the function ( f(x) = frac{2x^2 + sqrt{2} sinleft(x + frac{pi}{4}right)}{2x^2 + cos x} ), we need to find the maximum and minimum values ( a ) and ( b ) respectively, and then compute ( a + b ).First, we simplify the function using trigonometric identities. We note that:[ sinleft(x + frac{pi}{4}right) = frac{sin x + cos x}{sqrt{2}} ]Thus, the numerator becomes:[ 2x^2 + sqrt{2} sinleft(x + frac{pi}{4}right) = 2x^2 + sin x + cos x ]Therefore, the function can be rewritten as:[ f(x) = frac{2x^2 + sin x + cos x}{2x^2 + cos x} = 1 + frac{sin x}{2x^2 + cos x} ]Let ( g(x) = frac{sin x}{2x^2 + cos x} ). Then, ( f(x) = 1 + g(x) ).Next, we observe that ( g(x) ) is an odd function because:[ g(-x) = frac{sin(-x)}{2(-x)^2 + cos(-x)} = frac{-sin x}{2x^2 + cos x} = -g(x) ]Since ( g(x) ) is odd, its maximum value ( M ) and minimum value ( -M ) are symmetric around 0. Therefore, the maximum value of ( f(x) ) is ( 1 + M ) and the minimum value is ( 1 - M ). Adding these values together:[ a + b = (1 + M) + (1 - M) = 2 ]Thus, the final answer is:[ boxed{2} ],<think>Okay, so I need to find the maximum and minimum values of the function f(x) = (2x² + √2 sin(x + π/4)) / (2x² + cos x), and then add them together. Hmm, let's see. First, I should probably analyze the function to see how it behaves. Since both the numerator and the denominator involve 2x², which grows without bound as x approaches infinity or negative infinity, maybe the function has some horizontal asymptote. Let me check that.As x becomes very large in absolute value, the terms with x² dominate both the numerator and the denominator. So, the function would approach (2x²) / (2x²) = 1. So, the horizontal asymptote is y = 1. That means that as x goes to ±∞, f(x) approaches 1. Therefore, the function might oscillate around 1 but with diminishing amplitude. But since there are trigonometric functions in both numerator and denominator, the function could have local maxima and minima. The question is asking for the maximum and minimum values of the function over all real numbers, then adding them together.To find maxima and minima, the standard method is to take the derivative of f(x) and set it equal to zero. Let me recall that for a function f(x) = g(x)/h(x), the derivative is f’(x) = [g’(x)h(x) - g(x)h’(x)] / [h(x)]². So, I need to compute the derivatives of the numerator and the denominator first.Let me denote the numerator as N(x) = 2x² + √2 sin(x + π/4) and the denominator as D(x) = 2x² + cos x. Then, f(x) = N(x)/D(x).First, compute N’(x): derivative of 2x² is 4x, and derivative of √2 sin(x + π/4) is √2 cos(x + π/4). So, N’(x) = 4x + √2 cos(x + π/4).Then, D’(x): derivative of 2x² is 4x, and derivative of cos x is -sin x. So, D’(x) = 4x - sin x.Now, the derivative f’(x) is [N’(x)D(x) - N(x)D’(x)] / [D(x)]².So, let me write that out:f’(x) = [ (4x + √2 cos(x + π/4))(2x² + cos x) - (2x² + √2 sin(x + π/4))(4x - sin x) ] / (2x² + cos x)²This looks complicated, but maybe there's a way to simplify the numerator. Let's compute the numerator step by step.First, expand (4x + √2 cos(x + π/4))(2x² + cos x):= 4x * 2x² + 4x * cos x + √2 cos(x + π/4) * 2x² + √2 cos(x + π/4) * cos x= 8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos xThen, expand (2x² + √2 sin(x + π/4))(4x - sin x):= 2x² * 4x + 2x² * (-sin x) + √2 sin(x + π/4) * 4x + √2 sin(x + π/4) * (-sin x)= 8x³ - 2x² sin x + 4√2 x sin(x + π/4) - √2 sin(x + π/4) sin xNow, subtract the second expansion from the first:Numerator = [8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos x] - [8x³ - 2x² sin x + 4√2 x sin(x + π/4) - √2 sin(x + π/4) sin x]Let me subtract term by term:8x³ - 8x³ = 04x cos x - 0 = 4x cos x2√2 x² cos(x + π/4) - (-2x² sin x) = 2√2 x² cos(x + π/4) + 2x² sin x√2 cos(x + π/4) cos x - 0 = √2 cos(x + π/4) cos xThen, subtract the remaining terms:- [ -2x² sin x ] becomes +2x² sin x (wait, but I already accounted for that)Wait, maybe I should write the entire subtraction:Original first expansion: 8x³ + 4x cos x + 2√2 x² cos(x + π/4) + √2 cos(x + π/4) cos xMinus the second expansion: 8x³ - 2x² sin x + 4√2 x sin(x + π/4) - √2 sin(x + π/4) sin xSo:8x³ - 8x³ = 04x cos x - (-2x² sin x) ??? Wait, no. Wait, when subtracting the entire second expansion, each term is subtracted:So, 8x³ + 4x cos x + ... - [8x³ - 2x² sin x + ... ] = 8x³ -8x³ +4x cos x - (-2x² sin x) + ... etc.Wait, maybe I need to do this more carefully.First term: 8x³ - 8x³ = 0Second term: 4x cos x - (-2x² sin x) ??? Wait, no. Wait, the second expansion's first term is 8x³, so subtracting that gives -8x³. Then, the second term in the first expansion is 4x cos x, and the second term in the second expansion is -2x² sin x. Wait, actually, the order is different. Let's line them up properly:First expansion terms:1. 8x³2. +4x cos x3. +2√2 x² cos(x + π/4)4. +√2 cos(x + π/4) cos xSecond expansion terms:1. 8x³2. -2x² sin x3. +4√2 x sin(x + π/4)4. -√2 sin(x + π/4) sin xTherefore, subtracting the second expansion from the first:Term 1: 8x³ - 8x³ = 0Term 2: 4x cos x - (-2x² sin x) = 4x cos x + 2x² sin xTerm 3: 2√2 x² cos(x + π/4) - 4√2 x sin(x + π/4) = 2√2 x² cos(x + π/4) - 4√2 x sin(x + π/4)Term 4: √2 cos(x + π/4) cos x - (-√2 sin(x + π/4) sin x) = √2 [cos(x + π/4) cos x + sin(x + π/4) sin x]Therefore, combining all these:Numerator = 4x cos x + 2x² sin x + 2√2 x² cos(x + π/4) - 4√2 x sin(x + π/4) + √2 [cos(x + π/4) cos x + sin(x + π/4) sin x]Now, let's see if we can simplify this expression.First, note that cos(x + π/4) cos x + sin(x + π/4) sin x is a trigonometric identity. Remember that cos(A - B) = cos A cos B + sin A sin B. So here, A = x + π/4 and B = x, so cos((x + π/4) - x) = cos(π/4) = √2 / 2. Therefore, that term simplifies to √2 * (√2 / 2) = √2 * √2 / 2 = 2 / 2 = 1.So the last term simplifies to 1. So, the numerator becomes:4x cos x + 2x² sin x + 2√2 x² cos(x + π/4) - 4√2 x sin(x + π/4) + 1Hmm, okay. So now the numerator is:4x cos x + 2x² sin x + 2√2 x² cos(x + π/4) - 4√2 x sin(x + π/4) + 1This still looks quite complicated, but maybe we can factor some terms or use trigonometric identities. Let's see.First, note that x² terms: 2x² sin x + 2√2 x² cos(x + π/4). Let's factor out 2x²:2x² [ sin x + √2 cos(x + π/4) ]Similarly, the x terms: 4x cos x - 4√2 x sin(x + π/4). Factor out 4x:4x [ cos x - √2 sin(x + π/4) ]Then the remaining term is +1. So, the numerator can be written as:2x² [ sin x + √2 cos(x + π/4) ] + 4x [ cos x - √2 sin(x + π/4) ] + 1Hmm, maybe there's a way to simplify sin x + √2 cos(x + π/4). Let's compute that.First, expand cos(x + π/4):cos(x + π/4) = cos x cos π/4 - sin x sin π/4 = (√2 / 2)(cos x - sin x)Therefore, √2 cos(x + π/4) = √2*(√2 / 2)(cos x - sin x) = (2 / 2)(cos x - sin x) = cos x - sin xSo, sin x + √2 cos(x + π/4) = sin x + cos x - sin x = cos x. Oh, that's nice! So sin x + √2 cos(x + π/4) simplifies to cos x. That's a big simplification.Similarly, let's look at the other term: cos x - √2 sin(x + π/4). Let's compute that.Again, sin(x + π/4) = sin x cos π/4 + cos x sin π/4 = (√2 / 2)(sin x + cos x)Therefore, √2 sin(x + π/4) = √2*(√2 / 2)(sin x + cos x) = (2 / 2)(sin x + cos x) = sin x + cos xTherefore, cos x - √2 sin(x + π/4) = cos x - (sin x + cos x) = -sin xSo, that term simplifies to -sin x.Therefore, substituting back into the numerator:2x² [ cos x ] + 4x [ -sin x ] + 1 = 2x² cos x - 4x sin x + 1So, the numerator of f’(x) simplifies to 2x² cos x - 4x sin x + 1. That's a significant simplification. So, f’(x) = [2x² cos x - 4x sin x + 1] / (2x² + cos x)².Now, to find critical points, we need to set the numerator equal to zero:2x² cos x - 4x sin x + 1 = 0This equation seems still difficult to solve analytically. Maybe there's a substitution or particular value of x that satisfies this equation. Alternatively, since the problem is asking for maximum and minimum values, perhaps there is some symmetry or trigonometric identity that can be exploited.Alternatively, maybe we can evaluate f(x) at specific points where the derivative might be zero. Let's first check some simple points. For example, x = 0.At x = 0:Numerator N(0) = 0 + √2 sin(0 + π/4) = √2 * sin(π/4) = √2 * √2 / 2 = 1Denominator D(0) = 0 + cos 0 = 1So, f(0) = 1/1 = 1Now, check the derivative at x = 0:Numerator of f’(0): 0 - 0 + 1 = 1. Denominator is (0 + 1)^2 = 1. So f’(0) = 1/1 = 1. So, the derivative is positive at x = 0, meaning the function is increasing there.Hmm, maybe check x = π/4. Let's compute f(π/4):First, x = π/4.N(π/4) = 2*(π/4)^2 + √2 sin(π/4 + π/4) = 2*(π²/16) + √2 sin(π/2) = π²/8 + √2*1 ≈ (9.87)/8 + 1.414 ≈ 1.234 + 1.414 ≈ 2.648D(π/4) = 2*(π/4)^2 + cos(π/4) = π²/8 + √2/2 ≈ 1.234 + 0.707 ≈ 1.941Thus, f(π/4) ≈ 2.648 / 1.941 ≈ 1.364Now, check derivative at x = π/4. Wait, maybe this is getting messy. Alternatively, maybe we can consider that the function might attain its maximum and minimum at points where x is small, because as x grows, the function approaches 1, so the extrema might be near x = 0. Let's check x = π/2.At x = π/2:N(π/2) = 2*(π/2)^2 + √2 sin(π/2 + π/4) = 2*(π²/4) + √2 sin(3π/4) = (π²/2) + √2*(√2/2) = (π²/2) + 1 ≈ (9.87)/2 + 1 ≈ 4.935 + 1 = 5.935D(π/2) = 2*(π/2)^2 + cos(π/2) = same as numerator without the sin term, so 2*(π²/4) + 0 = π²/2 ≈ 4.935Thus, f(π/2) ≈ 5.935 / 4.935 ≈ 1.203So, f(π/2) ≈ 1.2. Hmm, less than at π/4. So maybe the maximum is around x = π/4? But that was approximately 1.364. Let's check x = π/3.Wait, maybe another approach. Since the derivative is complicated, maybe we can rewrite the original function in a different way to find its maximum and minimum. Let's look at f(x):f(x) = [2x² + √2 sin(x + π/4)] / [2x² + cos x]Note that the denominator is 2x² + cos x, which is always positive since 2x² is non-negative and cos x ≥ -1, so for |x| large enough, denominator is positive, and for small x, 2x² + cos x ≥ 0 + (-1) = -1, but actually, when x = 0, denominator is 1, and cos x is 1 at x=0. The minimum of denominator is when cos x is minimized, which is -1. So denominator is 2x² + cos x ≥ 2x² -1. Since 2x² -1 can be negative for |x| < 1/√2. So denominator could be negative for x between -1/√2 and 1/√2. Wait, but at x=0, it's 1. Hmm, maybe the denominator is always positive? Wait, cos x is between -1 and 1. So 2x² + cos x ≥ 2x² -1. For 2x² -1 to be positive, x² > 1/2, so |x| > 1/√2. For |x| ≤ 1/√2, 2x² ≤ 1, so 2x² + cos x ≥ 2x² -1. But when |x| ≤ 1/√2, 2x² ≤ 1, so 2x² -1 ≥ -1. So the denominator is ≥ -1 in that region, but actually, since cos x is at least -1, 2x² + cos x ≥ -1. However, since the denominator is in the function f(x), we need to make sure it doesn't equal zero. So, is there any x where 2x² + cos x = 0?Let's solve 2x² + cos x = 0. Since 2x² is always non-negative and cos x ≥ -1, 2x² + cos x ≥ -1. But 2x² + cos x = 0 would require 2x² = -cos x. Since left side is non-negative, right side must be non-positive. Therefore, -cos x ≥ 0 => cos x ≤ 0. So x in [π/2 + 2kπ, 3π/2 + 2kπ] for integer k. Then, 2x² = -cos x. But -cos x is between 0 and 1 in those intervals. So 2x² = -cos x => x² = (-cos x)/2. Since x² must be ≤ 1/2. So |x| ≤ 1/√2 ≈ 0.707. But in the intervals where cos x ≤ 0, x is at least π/2 ≈ 1.5708. So there is no real solution because |x| ≤ 1/√2 ≈ 0.707 and at the same time x must be ≥ π/2 ≈ 1.5708. Contradiction. Therefore, denominator 2x² + cos x is always positive. Therefore, the function f(x) is defined for all real numbers.Therefore, we don't have to worry about the denominator being zero. Good.Now, returning to the critical points. The equation to solve is 2x² cos x - 4x sin x + 1 = 0. This seems challenging. Maybe we can check if x=0 is a critical point. Plugging x=0 into the numerator: 0 -0 +1=1≠0. So, not a critical point. What about x=π/4?Let me compute 2*(π/4)^2 cos(π/4) -4*(π/4) sin(π/4) +1.First, (π/4)^2 ≈ (0.7854)^2 ≈ 0.6168cos(π/4) ≈ √2/2 ≈ 0.7071So first term: 2*0.6168*0.7071 ≈ 2*0.436 ≈ 0.872Second term: -4*(π/4)*sin(π/4) ≈ -π*0.7071 ≈ -2.221Third term: +1Total ≈ 0.872 -2.221 +1 ≈ -0.349. Not zero.How about x=π/2:2*(π/2)^2 cos(π/2) -4*(π/2) sin(π/2) +1 = 2*(π²/4)*0 -2π*1 +1 = 0 -2π +1 ≈ -6.283 +1 ≈ -5.283 ≠0x=π/3:x=π/3≈1.0472x² cos x ≈2*(1.047)^2*0.5≈2*1.096*0.5≈1.096-4x sin x≈-4*1.047*(√3/2)≈-4*1.047*0.866≈-4*0.906≈-3.624+1Total≈1.096 -3.624 +1≈-1.528≈-1.528≠0x=π/6:x≈0.52362x² cos x≈2*(0.274)*0.866≈2*0.237≈0.474-4x sin x≈-4*0.5236*0.5≈-4*0.2618≈-1.047+1Total≈0.474 -1.047 +1≈0.427≈0.427≠0Hmm, not zero. Maybe trying negative x.x=-π/4:2*(π/4)^2 cos(-π/4) -4*(-π/4) sin(-π/4) +1 = same as x=π/4 but with cos(-π/4)=cos(π/4)=√2/2, sin(-π/4)=-√2/2So:2*(π²/16)*(√2/2) -4*(-π/4)*(-√2/2) +1≈2*(0.6168)*(0.7071) - (-π)*( -0.7071 ) +1≈0.872 - (π*0.7071) +1≈0.872 -2.221 +1≈-0.349 same as before.Not zero.This is getting frustrating. Maybe there's a clever substitution or maybe the function has some symmetry. Let me look again at the original function.f(x) = [2x² + √2 sin(x + π/4)] / [2x² + cos x]Note that sin(x + π/4) can be written as sin x cos π/4 + cos x sin π/4 = (sin x + cos x)/√2. Therefore, √2 sin(x + π/4) = sin x + cos x. Therefore, the numerator simplifies:Numerator = 2x² + sin x + cos xDenominator = 2x² + cos xTherefore, f(x) = [2x² + sin x + cos x] / [2x² + cos x] = 1 + [sin x] / [2x² + cos x]Oh! That's a significant simplification. So f(x) = 1 + sin x / (2x² + cos x)So, since f(x) = 1 + sin x / D(x), where D(x) = 2x² + cos x. Therefore, to find the maximum and minimum of f(x), we need to find the maximum and minimum of sin x / D(x), and then add 1 to each.Therefore, if we let g(x) = sin x / (2x² + cos x), then f(x) = 1 + g(x). So, the maximum of f(x) is 1 + maximum of g(x), and the minimum of f(x) is 1 + minimum of g(x). Therefore, a + b = [1 + max g] + [1 + min g] = 2 + (max g + min g). So, if we can show that max g + min g = 0, then a + b = 2 + 0 = 2.Wait, is there a chance that g(x) is an odd function? Let's check. Let's see if g(-x) = -g(x).Compute g(-x) = sin(-x) / [2(-x)^2 + cos(-x)] = -sin x / [2x² + cos x] = -g(x). Yes! Because sin(-x) = -sin x, and cos(-x) = cos x, and 2x² is even. Therefore, g(-x) = -g(x). Therefore, g(x) is odd. Therefore, its graph is symmetric about the origin. Therefore, if (x, y) is on the graph, then (-x, -y) is also on the graph. Therefore, the maximum of g(x) is equal to the negative of the minimum of g(x). Therefore, max g = - min g. Therefore, max g + min g = 0. Therefore, a + b = 2 + 0 = 2.Therefore, the answer is 2.But wait, let me confirm this logic. If g is an odd function, then indeed its maximum and minimum are negatives of each other. So, if the maximum value of g is M, then the minimum value is -M, so their sum is 0. Therefore, f(x) has maximum 1 + M and minimum 1 - M, so a + b = (1 + M) + (1 - M) = 2. That's correct.Therefore, without even computing the derivative, using the fact that g(x) is odd, the sum of maximum and minimum of f(x) is 2.But wait, is this true? Let me check with the values I calculated earlier. For example, at x = π/4, f(x) ≈1.364, which is greater than 1. If the maximum is 1 + M and the minimum is 1 - M, then their average is 1. So, the function oscillates around 1, with amplitude M. Then, the maximum and minimum add to 2. But is there a case where this might not hold? For example, if the function has multiple maxima or minima, but since g is odd, every maximum at x corresponds to a minimum at -x. So, as long as the global maximum and global minimum are achieved at points x and -x, then their sum would be 2.But let's verify with the example values. When I calculated f(π/4) ≈1.364, so if that is a local maximum, then there should be a local minimum at x = -π/4. Let's check f(-π/4):Numerator: 2*(-π/4)^2 + √2 sin(-π/4 + π/4) = 2*(π²/16) + √2 sin(0) = π²/8 + 0 ≈1.234Denominator: 2*(-π/4)^2 + cos(-π/4) = π²/8 + √2/2 ≈1.234 +0.707≈1.941Therefore, f(-π/4) ≈1.234 /1.941≈0.636. Then, 0.636 is approximately 1 - 0.364, which would be the minimum. So, 1.364 + 0.636 = 2. So indeed, the sum is 2. So, this seems to hold in this case.Therefore, even without calculating the exact maxima and minima, using the symmetry of the function, we can conclude that a + b = 2.Final Answerboxed{2}</think>Given the function ( f(x) = frac{2x^2 + sqrt{2} sinleft(x + frac{pi}{4}right)}{2x^2 + cos x} ), we need to find the maximum and minimum values ( a ) and ( b ) respectively, and then compute ( a + b ).First, we simplify the function. By using the trigonometric identity ( sinleft(x + frac{pi}{4}right) = frac{sin x + cos x}{sqrt{2}} ), we rewrite the numerator:[sqrt{2} sinleft(x + frac{pi}{4}right) = sin x + cos x]Thus, the function becomes:[f(x) = frac{2x^2 + sin x + cos x}{2x^2 + cos x} = 1 + frac{sin x}{2x^2 + cos x}]Let ( g(x) = frac{sin x}{2x^2 + cos x} ). Then, ( f(x) = 1 + g(x) ).Next, we analyze the function ( g(x) ). Since ( g(-x) = frac{sin(-x)}{2(-x)^2 + cos(-x)} = frac{-sin x}{2x^2 + cos x} = -g(x) ), ( g(x) ) is an odd function. An odd function has the property that its maximum value is the negative of its minimum value. Therefore, if the maximum value of ( g(x) ) is ( M ), the minimum value is ( -M ).Consequently, the maximum value of ( f(x) ) is ( 1 + M ) and the minimum value is ( 1 - M ). Adding these values together:[a + b = (1 + M) + (1 - M) = 2]Thus, the final answer is:[boxed{2}]]

🔑:1. 2 Detailed Explanation: Convert the function to f(x)=frac{2 x^{2}+sqrt{2} sin left(x+frac{pi}{4}right)}{2 x^{2}+cos x}=1+frac{sin x}{2 x^{2}+cos x}. Let g(x)=frac{sin x}{2 x^{2}+cos x}, since it is an odd function, if there is a maximum value g(m), then there is a minimum value g(-m), so a+b=2,

🧠:[<think>Okay, let me try to solve these math problems one by one. I need to be careful with the order of operations, right? So, parentheses first, then multiplication and division from left to right, and finally addition and subtraction. Let me start with the first one.The first problem is 4.8×(3.5-2.1)÷7. Hmm, so inside the parentheses, 3.5 minus 2.1. Let me calculate that. 3.5 minus 2 is 1.5, and then subtract another 0.1, so that's 1.4. So now the expression becomes 4.8×1.4÷7. Next, I should do the multiplication and division from left to right. So first, 4.8 multiplied by 1.4. Let me compute that. 4 times 1.4 is 5.6, and 0.8 times 1.4 is 1.12. Adding those together: 5.6 + 1.12 equals 6.72. Now, divide that by 7. So 6.72 divided by 7. Let me see, 7 goes into 6.72. 7×0.9 is 6.3, subtract that from 6.72, we have 0.42. Then 7 goes into 0.42 exactly 0.06 times. So total is 0.96. Wait, 0.9 + 0.06 is 0.96. So the first answer is 0.96? Let me check that again. 4.8 times 1.4: 4.8*1 is 4.8, 4.8*0.4 is 1.92, so total is 4.8 + 1.92 = 6.72. Then 6.72 divided by 7. 7 into 6.72 is indeed 0.96. Okay, that seems right.Moving on to the second problem: 18.75 - 0.23×2 - 4.54. Here, order of operations says do multiplication before subtraction. So first calculate 0.23 times 2. That's 0.46. Then subtract that from 18.75. So 18.75 - 0.46 is 18.29. Then subtract 4.54 from that. 18.29 - 4.54. Let me compute this. 18.29 minus 4 is 14.29, minus 0.54 is 13.75. So the second answer is 13.75. Let me verify: 0.23×2=0.46. 18.75 - 0.46 = 18.29. 18.29 - 4.54: 18.29 minus 4.50 is 13.79, then plus 0.04 (since it's minus 0.54 which is 0.50 + 0.04). Wait, no. Wait, 18.29 - 4.54. Let me do it another way. Subtract 4 from 18.29: 14.29. Then subtract 0.54: 14.29 - 0.54. 14 minus 0.54 is 13.46, and then 0.29 - 0.54 is negative, so maybe I need to borrow. Wait, maybe I made a mistake here. Let me do it step by step. 18.29 minus 4.54. Let's write it as: 18.29- 4.54--------Starting from the right: 9 - 4 is 5. 2 - 5, can't do that, borrow from the left. So 2 becomes 12, and the 8 becomes 7. 12 - 5 = 7. Then 7 - 4 (from the 4.54) is 3. Then 1 - 0 is 1. Wait, but 18.29 minus 4.54 is 13.75? Wait, let's use another method. 4.54 is 4 + 0.54. So 18.29 - 4 = 14.29. Then 14.29 - 0.54. 14.29 minus 0.50 is 13.79, then minus 0.04 more is 13.75. Yes, that's correct. So 13.75 is the correct answer. Okay.Third problem: 0.9 + 99×0.9. Again, multiplication first. 99 times 0.9. Let's compute that. 100 times 0.9 is 90, minus 1 times 0.9 is 0.9, so 90 - 0.9 = 89.1. Then add 0.9 to that. 89.1 + 0.9 = 90. So the third answer is 90. Makes sense, because 0.9 is a common factor. So 0.9(1 + 99) = 0.9×100 = 90. Yep, that's a quicker way. Good.Fourth problem: 4 ÷ 0.8 - 0.8 ÷ 4. Need to do the divisions first, then subtraction. Let's compute each division. 4 divided by 0.8. 0.8 goes into 4 five times because 0.8×5 = 4. So that's 5. Then 0.8 divided by 4. 0.8 divided by 4 is 0.2. Then subtract 0.2 from 5: 5 - 0.2 = 4.8. Let me check again. 4/0.8: since 0.8 is 4/5, so 4 divided by (4/5) is 4*(5/4) = 5. Correct. Then 0.8/4 is 0.2. 5 - 0.2 is 4.8. So the fourth answer is 4.8.Wait, let me make sure I didn't mix up the order. The problem is 4 ÷ 0.8 minus 0.8 ÷ 4. Yes, so first term is 4 divided by 0.8, which is 5. Second term is 0.8 divided by 4, which is 0.2. Then subtract: 5 - 0.2 = 4.8. Correct.So all four answers:First: 0.96Second: 13.75Third: 90Fourth: 4.8I think that's all. Let me just quickly recap to make sure I didn't transpose any numbers or make any arithmetic errors. For the first problem, 3.5-2.1=1.4, then 4.8×1.4=6.72, divided by 7 is 0.96. Yep. Second problem, 0.23×2=0.46, 18.75-0.46=18.29, minus 4.54 gives 13.75. Third problem, 99×0.9=89.1 +0.9=90. Fourth problem, 4/0.8=5, 0.8/4=0.2, 5-0.2=4.8. All seems correct. Okay, confident with these answers.Final AnswerThe results are: 1. boxed{0.96} 2. boxed{13.75} 3. boxed{90} 4. boxed{4.8}</think>To solve the given problems, we follow the order of operations (parentheses, multiplication/division from left to right, addition/subtraction from left to right).1. Calculate (4.8 times (3.5 - 2.1) div 7): - First, compute the expression inside the parentheses: (3.5 - 2.1 = 1.4). - Then multiply: (4.8 times 1.4 = 6.72). - Finally, divide by 7: (6.72 div 7 = 0.96). - Result: (boxed{0.96}).2. Calculate (18.75 - 0.23 times 2 - 4.54): - First, perform the multiplication: (0.23 times 2 = 0.46). - Subtract this result from 18.75: (18.75 - 0.46 = 18.29). - Then subtract 4.54: (18.29 - 4.54 = 13.75). - Result: (boxed{13.75}).3. Calculate (0.9 + 99 times 0.9): - First, perform the multiplication: (99 times 0.9 = 89.1). - Then add 0.9: (89.1 + 0.9 = 90). - Result: (boxed{90}).4. Calculate (4 div 0.8 - 0.8 div 4): - First, perform the division: (4 div 0.8 = 5). - Then perform the other division: (0.8 div 4 = 0.2). - Subtract the second result from the first: (5 - 0.2 = 4.8). - Result: (boxed{4.8}).Final results:1. (boxed{0.96})2. (boxed{13.75})3. (boxed{90})4. (boxed{4.8}),<think>Okay, let me try to solve these math problems one by one. Hmm, starting with the first one: 4.8 multiplied by (3.5 minus 2.1) divided by 7. Alright, order of operations here, right? So parentheses first. Let me do 3.5 minus 2.1. Let me subtract those: 3.5 minus 2.1 is 1.4. Then the equation becomes 4.8 times 1.4 divided by 7. Now, multiplication and division are next, and they go from left to right. So first, 4.8 times 1.4. Let me calculate that. 4 times 1.4 is 5.6, and 0.8 times 1.4 is 1.12. Adding those together gives 5.6 plus 1.12 equals 6.72. Then divide that by 7. 6.72 divided by 7. Let's see, 7 goes into 6.72 how many times? 7 times 0.96 is 6.72, right? Because 7 times 0.9 is 6.3, and 7 times 0.06 is 0.42, so 6.3 plus 0.42 is 6.72. So the first answer is 0.96. Got that.Next problem: 18.75 minus 0.23 times 2 minus 4.54. Again, order of operations. Multiplication comes before subtraction. So first, 0.23 times 2. That's 0.46. Then the equation becomes 18.75 minus 0.46 minus 4.54. Let me compute that step by step. 18.75 minus 0.46 is... Hmm, 18.75 minus 0.4 is 18.35, then minus 0.06 more is 18.29. Then subtract 4.54 from 18.29. Let me do that. 18.29 minus 4 is 14.29, then minus 0.54 is 13.75. Wait, let me check again. 18.29 minus 4.54. 18.29 minus 4 is 14.29. Then subtract 0.54: 14.29 minus 0.5 is 13.79, minus 0.04 more is 13.75. Yeah, that seems right. So the second answer is 13.75.Third problem: 0.9 plus 99 times 0.9. Hmm, maybe there's a distributive property here. Let me see. 0.9 plus 99 times 0.9 is the same as 0.9 times (1 + 99). Because factoring out the 0.9. So 1 plus 99 is 100, so 0.9 times 100 is 90. That's much easier than doing 99 times 0.9 first. Let me confirm. 99 times 0.9 is 89.1, and then adding 0.9 gives 90. Yep, that works. So the third answer is 90.Last problem: 4 divided by 0.8 minus 0.8 divided by 4. So let's handle each division first. 4 divided by 0.8. Well, 0.8 times 5 is 4, so that's 5. Then 0.8 divided by 4. That's 0.2. Then subtract 0.2 from 5. 5 minus 0.2 is 4.8. Let me double-check. 4 divided by 0.8: 4 / 0.8 = 5. 0.8 divided by 4: 0.8 / 4 = 0.2. 5 - 0.2 = 4.8. Yes, that's correct. So the fourth answer is 4.8.I think that's all of them. Let me just go through each one again quickly to make sure I didn't make any arithmetic mistakes.First one: 3.5 - 2.1 = 1.4. 4.8 * 1.4 = 6.72. 6.72 / 7 = 0.96. Correct.Second one: 0.23 * 2 = 0.46. 18.75 - 0.46 = 18.29. 18.29 - 4.54 = 13.75. Right.Third one: 0.9 + 99*0.9 = 0.9*(1 + 99) = 0.9*100 = 90. Correct.Fourth one: 4/0.8 = 5, 0.8/4 = 0.2. 5 - 0.2 = 4.8. Yes.Alright, I think all the answers are correct.</think>Problem 1: 4.8 × (3.5 - 2.1) ÷ 7 First, compute the parentheses: 3.5 - 2.1 = 1.4 Then multiply and divide left to right: 4.8 × 1.4 = 6.72 6.72 ÷ 7 = 0.96 Answer: boxed{0.96}Problem 2: 18.75 - 0.23 × 2 - 4.54 Perform multiplication first: 0.23 × 2 = 0.46 Subtract sequentially: 18.75 - 0.46 = 18.29 18.29 - 4.54 = 13.75 Answer: boxed{13.75}Problem 3: 0.9 + 99 × 0.9 Factor out 0.9: 0.9 × (1 + 99) = 0.9 × 100 = 90 Answer: boxed{90}Problem 4: 4 ÷ 0.8 - 0.8 ÷ 4 Calculate each division: 4 ÷ 0.8 = 5 0.8 ÷ 4 = 0.2 Subtract the results: 5 - 0.2 = 4.8 Answer: boxed{4.8}]

🔑:Answer: (1) 4.8 times (3.5-2.1) div 7 = 4.8 times 1.4 div 7 = 4.8 times (1.4 div 7) = 4.8 times 0.2 = boxed{0.96} (2) 18.75 - 0.23 times 2 - 4.54 = 18.75 - 0.46 - 4.54 = 18.75 - (0.46 + 4.54) = 18.75 - 5 = boxed{13.75} (3) 0.9 + 99 times 0.9 = (1 + 99) times 0.9 = 100 times 0.9 = boxed{90} (4) 4 div 0.8 - 0.8 div 4 = 5 - 0.2 = boxed{4.8}

🧠:[<think>Okay, so I have this equation to solve: (x + 1998)(x + 1999)(x + 2000)(x + 2001) + 1 = 0. Hmm, that looks a bit intimidating at first glance with all those consecutive terms. Let me try to break it down step by step.First, I notice that the factors are four consecutive linear terms: (x + 1998), (x + 1999), (x + 2000), and (x + 2001). The numbers 1998 to 2001 are consecutive integers, so maybe there's a pattern here or a substitution that can simplify things.Let me think. If I let y be some expression related to x to make the terms symmetric. The middle points between the first and last terms might help. Let's see: The first term is x + 1998 and the last term is x + 2001. If I average these two, the midpoint would be at x + (1998 + 2001)/2 = x + 1999.5. Maybe substituting y = x + 1999.5 would center the terms around y.Let's try that substitution. Let y = x + 1999.5. Then:- x + 1998 = y - 1.5- x + 1999 = y - 0.5- x + 2000 = y + 0.5- x + 2001 = y + 1.5So the original equation becomes:(y - 1.5)(y - 0.5)(y + 0.5)(y + 1.5) + 1 = 0.Hmm, that looks a bit better. Now, this product of four terms with differences of 1 between them. Wait, actually, the terms are symmetric around y. Let me pair them up. Maybe multiply the first and last terms, and the middle two terms.So, (y - 1.5)(y + 1.5) * (y - 0.5)(y + 0.5) + 1 = 0.Calculating each pair:First pair: (y - 1.5)(y + 1.5) = y² - (1.5)² = y² - 2.25.Second pair: (y - 0.5)(y + 0.5) = y² - 0.25.So the equation becomes:(y² - 2.25)(y² - 0.25) + 1 = 0.Now, let's expand this product. Let me denote A = y². Then,(A - 2.25)(A - 0.25) + 1 = 0.Expanding the product:A² - 0.25A - 2.25A + (2.25)(0.25) + 1 = 0.Simplify the terms:A² - 2.5A + (0.5625) + 1 = 0.So, A² - 2.5A + 1.5625 = 0.Hmm, that's a quadratic in terms of A. Let's see if I can solve for A using the quadratic formula.Quadratic equation: A² - 2.5A + 1.5625 = 0.Discriminant D = (2.5)^2 - 4 * 1 * 1.5625 = 6.25 - 6.25 = 0.Oh, discriminant is zero, so there's one real root (a double root).Therefore, A = [2.5 ± sqrt(0)] / 2 = 2.5 / 2 = 1.25.So A = 1.25. Since A = y², that gives y² = 1.25.Therefore, y = ±√(1.25). Let me write that as √(5/4) = (√5)/2. So y = ±(√5)/2.Wait, 1.25 is 5/4, so yes, square root of 5/4 is √5 / 2. So y = ±√5 / 2.But remember that y = x + 1999.5, so substituting back:x + 1999.5 = ±√5 / 2.Therefore, solving for x:x = -1999.5 ± (√5)/2.So x = -1999.5 - (√5)/2 and x = -1999.5 + (√5)/2.But let me check if that's correct. Let me verify the steps again.Starting with substitution y = x + 1999.5, which shifts the terms to be symmetric. Then we rewrote the product as (y² - 2.25)(y² - 0.25) +1 =0, which became quadratic in y², leading to y² =1.25. So yes, y is ±√(1.25). Therefore x is -1999.5 ± √(5)/2.But wait, maybe there's another way to check.Alternatively, perhaps expanding the original equation. But that might be cumbersome. Let me see.Alternatively, notice that (x + 1998)(x + 2001) = x² + (1998 + 2001)x + 1998*2001 = x² + 3999x + 1998*2001.Similarly, (x + 1999)(x + 2000) = x² + (1999 + 2000)x + 1999*2000 = x² + 3999x + 1999*2000.So let me denote z = x² + 3999x + c, where c is some constant. Wait, but each product is x² + 3999x + different constants. Let me denote:Let me call the first product: (x + 1998)(x + 2001) = x² + 3999x + 1998*2001.Similarly, (x + 1999)(x + 2000) = x² + 3999x + 1999*2000.So let me set t = x² + 3999x + k, but maybe k is the average of 1998*2001 and 1999*2000? Let me compute those constants.First, 1998*2001. Let's compute that.1998*2001 = 1998*(2000 + 1) = 1998*2000 + 1998*1 = 3,996,000 + 1,998 = 3,997,998.Similarly, 1999*2000 = 1999*2000 = 3,998,000.So the two constants are 3,997,998 and 3,998,000. The average of these is (3,997,998 + 3,998,000)/2 = (7,995,998)/2 = 3,997,999. Hmm, not sure if that helps. Alternatively, maybe notice that 1998*2001 = (1999.5 - 1.5)(1999.5 + 1.5) = (1999.5)^2 - (1.5)^2, similar to the previous substitution. Wait, that's actually similar to what I did before.But since I already found the solution through substitution, maybe that's sufficient.Alternatively, let's check if the roots satisfy the original equation. Let's take x = -1999.5 + (√5)/2. Let's compute each term:x + 1998 = (-1999.5 + √5/2) + 1998 = -1.5 + √5/2.Similarly,x + 1999 = (-1999.5 + √5/2) + 1999 = -0.5 + √5/2.x + 2000 = (-1999.5 + √5/2) + 2000 = 0.5 + √5/2.x + 2001 = (-1999.5 + √5/2) + 2001 = 1.5 + √5/2.So the product is:(-1.5 + √5/2)(-0.5 + √5/2)(0.5 + √5/2)(1.5 + √5/2) + 1.Wait, but this seems complex. Let me compute it step by step.First, multiply (-1.5 + √5/2)(1.5 + √5/2) and (-0.5 + √5/2)(0.5 + √5/2).First pair: (-1.5 + √5/2)(1.5 + √5/2) = (-1.5)(1.5) + (-1.5)(√5/2) + (√5/2)(1.5) + (√5/2)(√5/2).Simplify:= -2.25 - (1.5√5)/2 + (1.5√5)/2 + (5)/4The middle terms cancel out, so:= -2.25 + 1.25 = -1.Similarly, the second pair: (-0.5 + √5/2)(0.5 + √5/2) = (-0.5)(0.5) + (-0.5)(√5/2) + (√5/2)(0.5) + (√5/2)(√5/2).Simplify:= -0.25 - (0.5√5)/2 + (0.5√5)/2 + (5)/4.Again, the middle terms cancel out:= -0.25 + 1.25 = 1.0.Therefore, the product is (-1)(1) = -1. Then adding 1 gives -1 + 1 = 0. So it checks out.Similarly, for the other root x = -1999.5 - √5/2, the product will be the same because the terms will be symmetric. Let me check:x + 1998 = -1999.5 - √5/2 + 1998 = -1.5 - √5/2.x + 1999 = -0.5 - √5/2.x + 2000 = 0.5 - √5/2.x + 2001 = 1.5 - √5/2.Multiply the first and last terms: (-1.5 - √5/2)(1.5 - √5/2).= (-1.5)(1.5) + (-1.5)(-√5/2) + (-√5/2)(1.5) + (-√5/2)(-√5/2)= -2.25 + (1.5√5)/2 - (1.5√5)/2 + (5)/4Again, middle terms cancel:= -2.25 + 1.25 = -1.Second pair: (-0.5 - √5/2)(0.5 - √5/2).= (-0.5)(0.5) + (-0.5)(-√5/2) + (-√5/2)(0.5) + (-√5/2)(-√5/2)= -0.25 + (0.5√5)/2 - (0.5√5)/2 + (5)/4Middle terms cancel:= -0.25 + 1.25 = 1.0.So again, product is (-1)(1) = -1, adding 1 gives 0. So both roots satisfy the equation.Therefore, the real roots are x = -1999.5 ± √5/2.But let me confirm that these are the only roots. The original equation is a quartic (degree 4) equation. However, after substitution, we found that the equation reduces to (y² - 2.25)(y² - 0.25) + 1 = 0, which led to a quadratic in y² with a double root. Therefore, y² = 1.25, so y has two real roots (positive and negative square roots), each giving a value of x. Thus, in total, two real roots. But quartic equations can have up to four real roots. However, in this case, due to the symmetry and the way the equation is constructed, we ended up with two real roots, each of multiplicity two? Wait, but the quadratic equation for A had a double root, so y² = 1.25, so each y gives two x's? Wait, no. Wait, y² = 1.25, so y = ±√(1.25), which is two values. Each y corresponds to one x. So two real roots. Hence, the quartic equation has two real roots, each with multiplicity two?Wait, but when we expanded (y² - 2.25)(y² - 0.25) +1 =0, we ended up with y^4 - 2.5y² + 1.5625 = 0, which factors as (y² - 1.25)^2 =0. Because (y² - 1.25)^2 = y^4 - 2.5y² + 1.5625. So indeed, the equation is (y² - 1.25)^2 =0. Therefore, y² =1.25, so y = ±√(1.25), each with multiplicity two. But since y is squared, each root y = √1.25 and y = -√1.25 would lead to one x each. Wait, no, the multiplicity would be two for each y, but since y is linear in x, each of these y roots would correspond to a single x. Therefore, the original equation would have two real roots, each with multiplicity two. However, when we checked, substituting x = -1999.5 ± √5/2 gives distinct roots, each with multiplicity one. Wait, maybe my initial analysis is wrong.Wait, let's see. The equation in terms of y is (y² - 1.25)^2 =0, which means that y² =1.25 has multiplicity two. Therefore, y can be √1.25 or -√1.25, each with multiplicity two. But since y is a first-degree substitution of x, each y value corresponds to exactly one x. Therefore, each of these roots for y would correspond to one x, but each x would have multiplicity two. However, when we checked, plugging in the x values into the original equation, they work. So perhaps the original equation has two real roots, each with multiplicity two. But let me check with another approach.Alternatively, suppose we consider the original quartic equation. Let me try to express it as a square plus 1 equals 0, which would imply that the square is -1, which is impossible unless the square itself is a negative number. Wait, but in our case, we had [(y² - 2.25)(y² - 0.25)] = -1. But through substitution, we found that this is possible when the product is -1. But through solving, we found that there are real roots, which makes sense because even though the product of four terms is negative, with the addition of 1 equalling zero, so the product is -1, which is possible if the terms are arranged such that their product is negative. But given that we found real solutions, it must be possible.Alternatively, perhaps the original equation can be written as a square. Let me see. If I consider the product (x + 1998)(x + 2001) = x² + (1998 + 2001)x + 1998*2001, which is x² + 3999x + 1998*2001. Similarly, (x + 1999)(x + 2000) = x² + 3999x + 1999*2000. Let me denote A = x² + 3999x + c, where c is a constant. If we let c be the average of 1998*2001 and 1999*2000, which we computed earlier as 3,997,999, but not sure if that helps.Alternatively, note that (x + 1998)(x + 1999)(x + 2000)(x + 2001) = [(x + 1998)(x + 2001)][(x + 1999)(x + 2000)] as done before, which equals [x² + 3999x + 1998*2001][x² + 3999x + 1999*2000]. Let me set z = x² + 3999x + k, where k is between 1998*2001 and 1999*2000. Let me compute the difference between these two constants: 1999*2000 - 1998*2001 = (1999)(2000) - (1998)(2001). Let's compute this:= 1999*2000 - 1998*2001= 1999*2000 - 1998*(2000 +1)= 1999*2000 - 1998*2000 - 1998*1= (1999 - 1998)*2000 - 1998= 1*2000 - 1998= 2000 - 1998= 2.So the two constants differ by 2. Let me denote:First constant: 1998*2001 = 3,997,998Second constant: 1999*2000 = 3,998,000So the product becomes (z + a)(z + b) where a = 3,997,998 and b = 3,998,000. Wait, but z = x² + 3999x, but no, z is x² + 3999x + k, but not sure. Alternatively, perhaps set z = x² + 3999x + 3,997,999.5 (the average). Then (z - 0.5)(z + 0.5) = z² - 0.25. Therefore, the product becomes z² - 0.25. Then, the original equation would be z² - 0.25 + 1 = z² + 0.75 =0, but that would lead to z² = -0.75, which has no real solutions. Wait, that doesn't align with previous findings. So maybe that approach is not correct.Wait, but earlier substitution gave real solutions, so there must be a mistake here. Let's check.Wait, z = x² + 3999x + 3,997,999.5Then, (z - 0.5)(z + 0.5) = z² - 0.25. But:Wait, if a = 3,997,998 and b = 3,998,000, then:(z - 0.5)(z + 0.5) = (x² + 3999x + 3,997,999.5 - 0.5)(x² + 3999x + 3,997,999.5 + 0.5) = (x² + 3999x + 3,997,999)(x² + 3999x + 3,998,000). But 3,997,999 is 1998*2001 +1, but not exactly. Wait, maybe this approach complicates things. Let's return to the substitution we had earlier which worked.So, the conclusion is that there are two real roots, each of multiplicity two, but when solving, we found each root once. Wait, but the equation in terms of y was (y² - 1.25)^2 =0, which would imply that y² =1.25 is a double root, so y = ±√1.25 each with multiplicity two. However, since y is linear in x, each y solution corresponds to a single x solution, hence each x solution would have multiplicity two. Therefore, the original equation would have two real roots, each with multiplicity two, leading to total four roots (with multiplicity). However, when we solved, we found two distinct real roots. Let me check the original equation's graph. A quartic with leading coefficient positive tends to positive infinity at both ends. The equation is [(x + a)(x + a+1)(x + a+2)(x + a+3)] +1 =0. Since the product of four linear terms is a quartic which is positive when x is very large or very negative, but adding 1 shifts it up by 1. Therefore, the equation (product) +1 =0 would have the product equal to -1. Since the product can be negative only if an odd number of the factors are negative. However, with four consecutive terms, depending on x, the product can be positive or negative. However, due to the symmetry, the minimum value of the product might be such that when shifted up by 1, it touches the x-axis at two points, each with multiplicity two. But according to our solution, it's touching at two distinct points each with multiplicity one. Wait, but the equation in y was (y² -1.25)^2 =0, which would suggest that when expanded, it has two roots each with multiplicity two. But in terms of x, each y root gives one x root. Therefore, the quartic equation would have two real roots, each with multiplicity two, hence total four roots (counting multiplicities). But when we solved, we found two distinct real roots, each of multiplicity one? Hmm, seems conflicting. Wait, perhaps there was a miscalculation.Wait, let's go back. When we substituted y = x + 1999.5, the equation became (y² - 2.25)(y² - 0.25) +1 =0. Let's expand this again:(y² - 2.25)(y² - 0.25) = y^4 - 0.25y² - 2.25y² + (2.25)(0.25) = y^4 - 2.5y² + 0.5625.Adding 1: y^4 - 2.5y² + 0.5625 +1 = y^4 - 2.5y² + 1.5625 =0.Factor this: Let's see if it factors into (y² + ay + b)(y² + cy + d). But since the equation is y^4 -2.5y² +1.5625 =0. Let me check if this is a perfect square. Let's see: Suppose it's (y² + p)^2. Then expanding: y^4 + 2py² + p². Comparing to y^4 -2.5y² +1.5625, we have 2p = -2.5 => p = -1.25, and p² = 1.5625. Which is exactly the case. Therefore, the equation is (y² -1.25)^2 =0. Therefore, the equation is a perfect square, so y² =1.25, with multiplicity two. Hence, y = ±√1.25 each with multiplicity two. Therefore, in terms of x, each y corresponds to a single x, so the original equation has two real roots, each with multiplicity two.But when we solved y² =1.25, we got y = ±√(5/4) = ±(√5)/2. Then x = -1999.5 ± (√5)/2. So two distinct real roots, each with multiplicity two. However, when we checked by plugging in, we only saw that they satisfy the equation, but multiplicities? If we factor the original equation, it would be [(x + 1999.5)^2 - (√5/2)^2]^2 =0, which is [(x + 1999.5 - √5/2)(x + 1999.5 + √5/2)]^2 =0. Hence, each linear factor squared. Therefore, the roots are x = -1999.5 ± √5/2, each with multiplicity two. So the quartic equation has two real roots, each repeated twice. Therefore, the answer should be these two roots, each with multiplicity two, but since the problem asks for all real roots, listing each once or twice?Wait, the problem says "Find all the real roots". In the context of competition problems, unless specified, they usually want the set of real roots, not considering multiplicity. So even if they have multiplicity two, we just list each distinct real root once. Therefore, the answer is x = -1999.5 ± (√5)/2. Which can be written as x = -1999.5 ± (√5)/2. Alternatively, expressing -1999.5 as -(1999 + 0.5) = -3999/2. So x = -3999/2 ± √5/2. Which combines to x = (-3999 ± √5)/2.Yes, that seems cleaner. Since -1999.5 is equal to -3999/2, then:x = -3999/2 + √5/2 = (-3999 + √5)/2,andx = -3999/2 - √5/2 = (-3999 - √5)/2.Therefore, the real roots are (-3999 ± √5)/2.Hence, the final answer is x = [-3999 ± √5]/2.Final AnswerThe real roots are boxed{dfrac{-3999 + sqrt{5}}{2}} and boxed{dfrac{-3999 - sqrt{5}}{2}}.</think>To find all the real roots of the equation ((x + 1998)(x + 1999)(x + 2000)(x + 2001) + 1 = 0), we start by making a substitution to simplify the expression. Let (y = x + 1999.5). This transforms the terms as follows:[begin{aligned}x + 1998 &= y - 1.5, x + 1999 &= y - 0.5, x + 2000 &= y + 0.5, x + 2001 &= y + 1.5.end{aligned}]The equation then becomes:[(y - 1.5)(y - 0.5)(y + 0.5)(y + 1.5) + 1 = 0.]We pair the terms and simplify:[begin{aligned}(y - 1.5)(y + 1.5) &= y^2 - (1.5)^2 = y^2 - 2.25, (y - 0.5)(y + 0.5) &= y^2 - (0.5)^2 = y^2 - 0.25.end{aligned}]Substituting these results back into the equation, we get:[(y^2 - 2.25)(y^2 - 0.25) + 1 = 0.]Expanding the product:[(y^2 - 2.25)(y^2 - 0.25) = y^4 - 2.5y^2 + 0.5625.]Adding 1 to both sides:[y^4 - 2.5y^2 + 1.5625 = 0.]This can be factored as a perfect square:[(y^2 - 1.25)^2 = 0.]Solving for (y), we get:[y^2 = 1.25 implies y = pm frac{sqrt{5}}{2}.]Substituting back (y = x + 1999.5), we find:[x + 1999.5 = pm frac{sqrt{5}}{2}.]Solving for (x):[x = -1999.5 pm frac{sqrt{5}}{2}.]Expressing (-1999.5) as (-frac{3999}{2}), the solutions are:[x = frac{-3999 pm sqrt{5}}{2}.]Thus, the real roots are:[boxed{dfrac{-3999 + sqrt{5}}{2}} quad text{and} quad boxed{dfrac{-3999 - sqrt{5}}{2}}.],<think>Alright, so I need to find all the real roots of the equation ((x + 1998)(x + 1999)(x + 2000)(x + 2001) + 1 = 0). Hmm, okay. Let's start by understanding what this equation is. It's a product of four consecutive linear terms plus 1 equals zero. That seems like it might be a quartic equation, which can be tricky, but maybe there's a way to simplify it.First, I notice that the terms are consecutive: 1998, 1999, 2000, 2001. So, they form an arithmetic sequence with a common difference of 1. Maybe I can make a substitution to center the terms around a middle value. Let's see, the middle points between 1998 and 2001 would be at 1999.5. Let me check: 1998 is 1999.5 - 1.5, 1999 is 1999.5 - 0.5, 2000 is 1999.5 + 0.5, and 2001 is 1999.5 + 1.5. So, if I let ( y = x + 1999.5 ), that might help simplify the expression.Let me try that substitution. Let ( y = x + 1999.5 ). Then, each term becomes:- ( x + 1998 = y - 1.5 )- ( x + 1999 = y - 0.5 )- ( x + 2000 = y + 0.5 )- ( x + 2001 = y + 1.5 )So, substituting into the original equation:[(y - 1.5)(y - 0.5)(y + 0.5)(y + 1.5) + 1 = 0]Hmm, okay. Now, this product looks like it could pair terms in a way that might form a difference of squares. Let me group the first two and the last two terms:[[(y - 1.5)(y + 1.5)][(y - 0.5)(y + 0.5)] + 1 = 0]Calculating each pair:First pair: ( (y - 1.5)(y + 1.5) = y^2 - (1.5)^2 = y^2 - 2.25 )Second pair: ( (y - 0.5)(y + 0.5) = y^2 - (0.5)^2 = y^2 - 0.25 )So now, the equation becomes:[(y^2 - 2.25)(y^2 - 0.25) + 1 = 0]Let me multiply those two quadratic terms. Let's let ( z = y^2 ), so the equation is:[(z - 2.25)(z - 0.25) + 1 = 0]Expanding the product:[z^2 - 0.25z - 2.25z + (2.25)(0.25) + 1 = 0Simplify the terms:First, combine the z terms: ( -0.25z - 2.25z = -2.5z )Then, calculate the constant term: ( (2.25)(0.25) = 0.5625 )So, the equation becomes:[z^2 - 2.5z + 0.5625 + 1 = 0Combine the constants: 0.5625 + 1 = 1.5625So, equation is:[z^2 - 2.5z + 1.5625 = 0Hmm, this is a quadratic in z. Let's try to solve it using the quadratic formula. The quadratic formula is ( z = frac{2.5 pm sqrt{(2.5)^2 - 4 cdot 1 cdot 1.5625}}{2} )Calculating discriminant:( (2.5)^2 = 6.25 )( 4 cdot 1 cdot 1.5625 = 6.25 )So, discriminant is ( 6.25 - 6.25 = 0 )Therefore, there is a repeated root at ( z = frac{2.5}{2} = 1.25 )So, z = 1.25But z = y^2, so y^2 = 1.25Therefore, y = ±√1.25Simplify √1.25: √(5/4) = (√5)/2 ≈ 1.118, but exact form is (√5)/2.Therefore, y = (√5)/2 or y = -√5/2Now, recall that y = x + 1999.5Therefore, solving for x:x = y - 1999.5So, the two solutions are:x = (√5)/2 - 1999.5andx = -√5/2 - 1999.5But wait, the original equation is a quartic, which should have four roots (though some might be complex). But we have only two real roots here. Let me check if I made a mistake.Wait, when we made the substitution z = y^2, and found that z = 1.25, so y^2 = 1.25 gives y = ±√1.25. Each of these y values would correspond to a real x value. So, two real roots. But a quartic with real coefficients can have 0, 2, or 4 real roots. The question is asking for all real roots, so maybe there are only two?Wait, but let me verify by plugging back into the original equation. Let me check.But first, let's see if the substitution and the steps are correct.Original equation:(x + 1998)(x + 1999)(x + 2000)(x + 2001) + 1 = 0Substituted y = x + 1999.5, so shifted the variable to center the terms.Then, paired (y - 1.5)(y + 1.5) and (y - 0.5)(y + 0.5) which are differences of squares. That's correct.Then multiplied (y² - 2.25)(y² - 0.25) + 1 = 0. Then set z = y², leading to (z - 2.25)(z - 0.25) + 1 = 0. Then expanded to z² - 2.5z + 0.5625 + 1 = z² - 2.5z + 1.5625 = 0. Then discriminant was 0, leading to z = 1.25. Then y² = 1.25, so y = ±√5/2, which is approximately ±1.118. Then x = y - 1999.5.So, x = -1999.5 ± √5/2. Therefore, two real roots. But quartic equation, so maybe the other two roots are complex?Let me check if the original equation can be written as a square. Since the quadratic in z had a discriminant of 0, so (z - 1.25)^2 = 0. Therefore, going back:(z - 1.25)^2 = 0 => (y² - 1.25)^2 = 0 => y² = 1.25. So, multiplicity 2 for each y solution. Wait, no. When we have (y² - 1.25)^2 = 0, then y² = 1.25, so y = ±√1.25, each with multiplicity 2. But since y² = 1.25 has two solutions, each of multiplicity 1, but because the original equation is (y² - 1.25)^2 = 0, so each root is of multiplicity 2. Wait, maybe not. Let me check.Wait, when we had (y² - 2.25)(y² - 0.25) + 1 = 0, and then we found that equals to (y² - 1.25)^2 = 0. Wait, let me verify that.Wait, (z - 2.25)(z - 0.25) + 1 = z² - 2.5z + 0.5625 + 1 = z² - 2.5z + 1.5625. Then, z² - 2.5z + 1.5625 = (z - 1.25)^2, because (z - 1.25)^2 = z² - 2.5z + 1.5625. Yes, that's correct. Therefore, the equation is (y² - 1.25)^2 = 0. Therefore, y² = 1.25 with multiplicity 2, leading to y = ±√1.25, each with multiplicity 2. But since y is a real variable, each simple root y = √1.25 and y = -√1.25 would correspond to two real roots each, but since they are squared, perhaps each leads to one real root? Wait, no. Wait, if we have y² = 1.25, then y = ±√1.25, so two distinct real solutions. Each with multiplicity 1. Wait, but the equation (y² - 1.25)^2 = 0 would imply that y² - 1.25 is squared, so the multiplicity is 2. But when solving y² = 1.25, the roots y = ±√1.25 each have multiplicity 1 in the equation y² = 1.25, but since it's squared in the original equation, does that mean each root y has multiplicity 2? Hmm, perhaps.Wait, the original equation after substitution is (y² - 1.25)^2 = 0. So, that is the same as (y - √1.25)^2 (y + √1.25)^2 = 0. Therefore, each root y = √1.25 and y = -√1.25 has multiplicity 2. But when solving for x, each y would give one x. Therefore, x = -1999.5 + √1.25 and x = -1999.5 - √1.25, each with multiplicity 2. Therefore, even though the equation is quartic, there are two distinct real roots, each with multiplicity 2. So, the real roots are x = -1999.5 ± √5/2, each repeated twice. But the problem asks for all the real roots, so we can list them as x = -1999.5 + √5/2 and x = -1999.5 - √5/2. Since multiplicities don't affect the set of roots, just their count.Therefore, the real roots are x = -1999.5 ± (√5)/2.But let me verify this by plugging back into the original equation. Let's take x = -1999.5 + (√5)/2.Compute x + 1998: (-1999.5 + √5/2) + 1998 = -1.5 + √5/2Similarly, x + 1999 = -0.5 + √5/2x + 2000 = 0.5 + √5/2x + 2001 = 1.5 + √5/2So, the product is (-1.5 + √5/2)(-0.5 + √5/2)(0.5 + √5/2)(1.5 + √5/2)Let me compute this step by step.First, pair the terms as before:[(-1.5 + √5/2)(1.5 + √5/2)] and [(-0.5 + √5/2)(0.5 + √5/2)]Compute first pair:(-1.5 + √5/2)(1.5 + √5/2) = (√5/2)^2 - (1.5)^2 = (5/4) - (2.25) = 1.25 - 2.25 = -1Second pair:(-0.5 + √5/2)(0.5 + √5/2) = (√5/2)^2 - (0.5)^2 = 5/4 - 0.25 = 1.25 - 0.25 = 1Multiply the two results: (-1)(1) = -1Then add 1: -1 + 1 = 0, which matches the equation. So that works.Similarly, check x = -1999.5 - √5/2.x + 1998 = -1.5 - √5/2x + 1999 = -0.5 - √5/2x + 2000 = 0.5 - √5/2x + 2001 = 1.5 - √5/2Again, pair terms:[(-1.5 - √5/2)(1.5 - √5/2)] and [(-0.5 - √5/2)(0.5 - √5/2)]First pair:(-1.5 - √5/2)(1.5 - √5/2) = (-1.5)(1.5) + (-1.5)(-√5/2) + (-√5/2)(1.5) + (-√5/2)(-√5/2)But maybe easier to recognize as (a - b)(a + b) where a = -1.5, b = √5/2? Wait, no. Wait, let's compute it as follows:Multiply (-1.5 - √5/2)(1.5 - √5/2) = (-1.5)(1.5) + (-1.5)(-√5/2) + (-√5/2)(1.5) + (-√5/2)(-√5/2)Calculates to:-2.25 + (1.5√5)/2 - (1.5√5)/2 + (5)/4Simplify:The middle terms cancel: (1.5√5)/2 - (1.5√5)/2 = 0So left with -2.25 + 1.25 = -1Similarly, second pair:[(-0.5 - √5/2)(0.5 - √5/2)] = (-0.5)(0.5) + (-0.5)(-√5/2) + (-√5/2)(0.5) + (-√5/2)(-√5/2)Calculates to:-0.25 + (0.5√5)/2 - (0.5√5)/2 + (5)/4Again, middle terms cancel: (0.5√5)/2 - (0.5√5)/2 = 0Left with -0.25 + 1.25 = 1Multiply the two results: (-1)(1) = -1, add 1 gives 0. So that works too.Therefore, both roots satisfy the equation. Hence, the real roots are x = -1999.5 ± √5/2.But let me confirm if there are any other real roots. The equation is a quartic, so degree 4. If we have two real roots each with multiplicity 2, that accounts for all four roots (with multiplicity). But if the quartic can be factored into a square of a quadratic, then it's possible that the other roots are complex. Let's check.We had the equation transformed into (y² - 1.25)^2 = 0, which factors as (y - √1.25)^2 (y + √1.25)^2 = 0. Therefore, each of these roots is a double root. So in terms of x, each x = -1999.5 ± √1.25 is a double root. Therefore, the equation has two distinct real roots, each with multiplicity 2. So, when the problem asks for all real roots, it's just these two, each repeated twice. But since they just want the roots, not considering multiplicity, the answer is those two values.Alternatively, if they want the roots with multiplicity, but the problem says "all the real roots," so likely just the two distinct ones.Thus, the real roots are x = -1999.5 + (√5)/2 and x = -1999.5 - (√5)/2.But √1.25 is equal to √(5/4) which is √5 / 2, so yes, that's correct.Therefore, the answer should be those two expressions. Let me write them in a boxed format.Final AnswerThe real roots are (boxed{-1999.5 + dfrac{sqrt{5}}{2}}) and (boxed{-1999.5 - dfrac{sqrt{5}}{2}}).Alternatively, since -1999.5 is equal to -(dfrac{3999}{2}), we could write it as:(boxed{-dfrac{3999}{2} + dfrac{sqrt{5}}{2}}) and (boxed{-dfrac{3999}{2} - dfrac{sqrt{5}}{2}})But the problem might prefer decimal or another form? Wait, the original problem didn't specify, but since 1999.5 is a fractional number, but expressed as a decimal. However, in mathematical problems, fractions are often preferred unless decimals are specified. However, 1999.5 is equal to (frac{3999}{2}). So perhaps writing it as (-frac{3999}{2} pm frac{sqrt{5}}{2}). Combining the terms, that would be (frac{-3999 pm sqrt{5}}{2}). Which is a more compact form.Yes, combining the terms:x = (-3999 ± √5)/2So, written as:boxed{dfrac{-3999 + sqrt{5}}{2}} and boxed{dfrac{-3999 - sqrt{5}}{2}}But in the problem statement, they might expect both roots in one box, but the instructions say "put your final answer within boxed{}". Since there are two roots, maybe we need to present both in separate boxes or combined. Looking at the problem statement example, sometimes multiple roots are given in separate boxes. However, in the original instruction, the user says "put your final answer within boxed{}", so perhaps they expect both answers in one box, separated by commas.But the standard practice in many Olympiad problems is to present all roots in a single box, separated by commas if there are multiple. Let me check the problem statement again: "Find all the real roots...". So, since there are two real roots, we need to present both.Therefore, the final answer is:boxed{-dfrac{3999}{2} + dfrac{sqrt{5}}{2}} quad text{and} quad boxed{-dfrac{3999}{2} - dfrac{sqrt{5}}{2}}But given that the user might expect both answers in one line. Alternatively, since the roots can be written as (frac{-3999 pm sqrt{5}}{2}), which is a more compact form. So:boxed{dfrac{-3999 pm sqrt{5}}{2}}But the ± symbol is acceptable in boxed answers, so this might be preferable.Yes, that's a more concise way. So, writing both roots together using ±.Final Answerboxed{dfrac{-3999 pm sqrt{5}}{2}}</think>To find all the real roots of the equation ((x + 1998)(x + 1999)(x + 2000)(x + 2001) + 1 = 0), we start by making a substitution to simplify the expression. Let ( y = x + 1999.5 ). This transforms the terms as follows:- ( x + 1998 = y - 1.5 )- ( x + 1999 = y - 0.5 )- ( x + 2000 = y + 0.5 )- ( x + 2001 = y + 1.5 )Substituting these into the original equation, we get:[(y - 1.5)(y - 0.5)(y + 0.5)(y + 1.5) + 1 = 0]Grouping the terms into pairs and using the difference of squares, we have:[[(y - 1.5)(y + 1.5)][(y - 0.5)(y + 0.5)] + 1 = 0]Calculating each pair:[(y^2 - 2.25)(y^2 - 0.25) + 1 = 0]Let ( z = y^2 ). The equation becomes:[(z - 2.25)(z - 0.25) + 1 = 0]Expanding and simplifying:[z^2 - 2.5z + 1.5625 = 0]This quadratic equation has a discriminant of 0, leading to a repeated root ( z = 1.25 ). Thus, ( y^2 = 1.25 ), giving ( y = pm sqrt{1.25} = pm frac{sqrt{5}}{2} ).Substituting back to find ( x ):[x = y - 1999.5]Thus, the real roots are:[x = -1999.5 + frac{sqrt{5}}{2} quad text{and} quad x = -1999.5 - frac{sqrt{5}}{2}]Expressing these in a more compact form, we get:[boxed{dfrac{-3999 pm sqrt{5}}{2}}]]

🔑:x = -3999/2 ± ½ √5 Solution Put y = x 2 + 3999x + 1998·2001. Then the polynomial is y(y + 2) + 1 = (y + 1) 2 . Thanks to Vikash Rungta for sending this solution. 19th Aus 1998 © John Scholes [email protected] 1 Mar 2003 Last updated/corrected 1 Mar 03

🧠:[<think>Okay, so the problem is asking how many zeros the product of all integers from 1 to 100 ends with. That product is 100 factorial, written as 100!. Hmm, I remember that the number of trailing zeros in a factorial is related to how many times the number 10 is a factor in that product. Since 10 is 2 multiplied by 5, the number of trailing zeros depends on the number of pairs of 2s and 5s in the prime factorization of 100!. But wait, I also recall that there are usually more 2s than 5s in the prime factors of factorials, so the number of 5s is the limiting factor here. So the key is to count how many times 5 appears as a factor in the numbers from 1 to 100.Let me think. To find the number of times 5 is a factor in 100!, you can use the formula where you divide 100 by 5, then divide by 25, then by 125, and so on, taking the floor of each division and adding them up. Since 5 squared is 25, 5 cubed is 125, etc. Each of these accounts for numbers that contribute more than one 5. For example, 25 is 5*5, so it contributes two 5s. Similarly, 50 is 2*5*5, contributing two 5s, and 75 is 3*5*5, also two 5s. Then numbers like 125 would contribute three 5s, but since we're only going up to 100, 125 is beyond that. So we only need to consider 5 and 25 here.Let me calculate. First, divide 100 by 5. 100 ÷ 5 = 20. So there are 20 multiples of 5 between 1 and 100. Then divide 100 by 25. 100 ÷ 25 = 4. So there are 4 multiples of 25. Each of these contributes an extra 5. Then divide 100 by 125. 100 ÷ 125 = 0.8, but we take the floor which is 0. So we can stop here. Then total number of 5s is 20 + 4 = 24.But let me verify that. For example, the multiples of 5 between 1 and 100 are 5, 10, 15, 20, 25, ..., 100. That's 20 numbers. Each contributes at least one 5. Then among these, the multiples of 25 (which is 25, 50, 75, 100) contribute an additional 5 each. So that's 4 more 5s. So total 20 + 4 = 24. Therefore, there are 24 5s in the prime factorization of 100!.Since there are more 2s than 5s, the number of trailing zeros is determined by the number of 5s. Therefore, the answer should be 24. But wait, let me check if I missed anything. For example, 100 divided by 5 is 20, divided by 25 is 4, divided by 125 is 0. So 20 + 4 =24. Yes, that seems correct. Let me cross-validate. For example, 25 is 5*5, so each multiple of 25 gives an extra 5. Then 50 is 2*5*5, so same. 75 is 3*5*5, and 100 is 4*5*5. So each of these four numbers (25,50,75,100) gives an additional 5. So indeed, 4 extra 5s. Then the rest of the multiples of 5 (20 total) give one 5 each, but since 4 of them are already counted in the 25 division, does that mean I need to adjust? Wait, no, because when I divide by 5, I get all multiples of 5, including those that are multiples of 25. Then dividing by 25 gives the additional 5s for those. So the formula is correct because each multiple of 25 contributes two 5s, so the first division by 5 counts one 5 for each multiple of 25, and the division by 25 counts the second 5. Similarly, if there were multiples of 125, they would contribute three 5s, so they'd be counted once in division by 5, once in division by 25, and once in division by 125, summing up to three. Since 125 is over 100, we don't have that here.Therefore, the calculation seems correct. 20 + 4 =24. Therefore, the number of trailing zeros in 100! is 24. Let me see if another way of thinking gives the same result. For example, manually checking smaller factorials. Let's take 25!.For 25!, number of 5s is 25 ÷5 =5, 25 ÷25=1, so 5+1=6. Then 25! has 6 trailing zeros. Let's check 25! is 15511210043330985984000000. Wait, no, that's 25 factorial? Maybe I should calculate 25! /10^6 and see if it's an integer. 25! = 15511210043330985984000000. Wait, actually, 25! is 15511210043330985984000000, which has 6 trailing zeros. Wait, but 25 divided by 5 is 5, divided by 25 is 1, so 5 +1 =6. Correct. Then 100! would have 24. So that seems consistent.Alternatively, let's take 10!. 10/5=2, 10/25=0, so 2 trailing zeros. 10! is 3628800, which has two zeros. Correct. 25! has 6 zeros. So the formula works. Therefore, applying it to 100! gives 24. So I think the answer is 24.Final AnswerThe product of all integers from 1 to 100 ends with boxed{24} zeros.</think>To determine how many zeros the product of all integers from 1 to 100 ends with, we need to count the number of times 10 is a factor in the product. Since 10 is the product of 2 and 5, and there are usually more factors of 2 than 5 in a factorial, the number of trailing zeros is determined by the number of times 5 is a factor.We use the formula to count the number of 5s in the prime factorization of 100! by dividing 100 by powers of 5 and taking the floor of each division:1. Divide 100 by 5: ( leftlfloor frac{100}{5} rightrfloor = 20 )2. Divide 100 by 25: ( leftlfloor frac{100}{25} rightrfloor = 4 )3. Divide 100 by 125: ( leftlfloor frac{100}{125} rightrfloor = 0 )Adding these results gives the total number of 5s: ( 20 + 4 = 24 ).Thus, the number of trailing zeros in 100! is (boxed{24}).,<think>Okay, so I need to figure out how many zeros the product of all integers from 1 to 100 ends with. Hmm, that's 100 factorial, right? Which is written as 100! So, the question is asking for the number of trailing zeros in 100!.Wait, trailing zeros... I remember that each trailing zero comes from multiplying 10, which is 2 times 5. And since there are usually more 2s than 5s in the prime factors of a factorial, the number of trailing zeros should be determined by the number of 5s in the prime factorization of 100!. Is that right?Let me verify. For example, take 10!. The prime factors include multiples of 2 and 5. The multiples of 2 are 2, 4, 6, 8, 10. The multiples of 5 are 5, 10. So there are two 5s and five 2s. Thus, 10! should end with two zeros. And indeed, 10! is 3,628,800, which has two zeros. So that checks out.Therefore, the number of trailing zeros in 100! is equal to the number of times 5 is a factor in the numbers from 1 to 100. But how do I calculate that?Well, every multiple of 5 contributes at least one 5. But some numbers contribute more than one 5, like 25 is 5 squared, so it contributes two 5s. Similarly, 50, 75, 100 are multiples of 25 and contribute two 5s each. And 125 would contribute three, but 125 is beyond 100, so we don't have to worry about that here.So the formula, if I recall correctly, is to divide 100 by 5, take the floor of that, then divide 100 by 25, take the floor, then divide by 125, and so on until the division result is zero. Then sum all those up.Let me try that. So first, 100 divided by 5 is 20. Then 100 divided by 25 is 4. Then 100 divided by 125 is 0.8, but we take the floor, so 0. So adding 20 + 4 + 0 = 24. Therefore, there should be 24 trailing zeros in 100!.Wait, let me make sure I'm doing that correctly. Each multiple of 5 gives one 5, each multiple of 25 gives an extra 5, each multiple of 125 gives another, etc. So for numbers up to 100, multiples of 5: 20 numbers (5, 10, 15, ..., 100). Multiples of 25: 4 numbers (25, 50, 75, 100). Multiples of 125: 0, since 125 > 100. So total 20 + 4 = 24. That seems correct.To check, maybe I can do a smaller factorial where I can compute the number of zeros. Let's take 25!.Number of trailing zeros would be 25/5 = 5, 25/25 = 1, 25/125 = 0. So 5 + 1 = 6 zeros. Let's compute 25!.But 25! is a huge number. Alternatively, maybe I can check a known value. For example, 25! is known to have 6 trailing zeros. Let me confirm with a calculator. 25! is 15511210043330985984000000. Wait, that's 15 zeros? Wait, that can't be right. Hmm, maybe I did something wrong.Wait, actually, 25! is 15511210043330985984000000. Let me count the zeros at the end. Starting from the right: 0, 0, 0, 0, 0, 0, 0? Wait, seven zeros? Hmm, that contradicts the calculation. Wait, maybe my example is wrong. Alternatively, maybe the calculator I'm using is incorrect.Wait, perhaps I should not rely on that. Let me try another way. Let's compute 25! step by step.25! = 25 × 24 × ... × 1. Let's track the number of factors of 5. Let's list all numbers from 1 to 25 and count the number of 5s.Numbers divisible by 5: 5, 10, 15, 20, 25. That's five numbers. Each contributes at least one 5. Additionally, 25 is 5², so it contributes an extra 5. So total number of 5s is 5 + 1 = 6. So 25! should have 6 trailing zeros. Therefore, if my calculator shows more, perhaps the calculator is incorrect or I'm miscounting.Alternatively, perhaps I made a mistake in looking up 25!.Alternatively, maybe the formula is correct. Let me check 10! again. 10/5 = 2, 10/25 = 0. So 2 zeros, which matches 3628800, which has two trailing zeros. So that works.Another example: 30!.30/5 = 6, 30/25 = 1, 30/125 = 0. So total 7 trailing zeros. Let me check 30!.30! is 265252859812191058636308480000000. Let's see, the trailing zeros: from the end, how many? Let's count: 0,0,0,0,0,0,0. So seven zeros. That's correct.Therefore, the formula works here. So back to 100!, the number of trailing zeros is 20 (from 100/5) + 4 (from 100/25) + 0 (from 100/125) = 24. So the answer should be 24.Wait, but just to be thorough, let me step through all numbers from 1 to 100 and count the number of 5s.Every fifth number is a multiple of 5, so 100 /5 = 20 numbers. But some numbers are multiples of 25, which is 5x5, so they contribute an extra 5. How many multiples of 25 are there? 100 /25 = 4. Then multiples of 125, but 125 >100, so none. So total 20 +4=24. That seems right.Alternatively, maybe I can list them:Multiples of 5: 5,10,15,20,25,30,...,100. That's 20 numbers.Multiples of 25: 25,50,75,100. That's 4 numbers. Each of these contributes an additional 5.So total 5s: 20 +4=24. Since there are plenty of 2s, the number of 10s is determined by the number of 5s, which is 24. Therefore, 24 trailing zeros.I think that's solid. So the answer is 24.Final AnswerThe product of all integers from 1 to 100 ends with boxed{24} zeros.</think>To determine how many zeros the product of all integers from 1 to 100 ends with, we need to count the number of trailing zeros in 100!. Trailing zeros are formed by the product of 10, which is 2 × 5. Since there are usually more factors of 2 than 5 in a factorial, the number of trailing zeros is determined by the number of factors of 5.We calculate the number of factors of 5 in 100! by considering the multiples of 5, 25, 125, etc., and summing the quotients of 100 divided by each of these numbers:1. Multiples of 5: ( leftlfloor frac{100}{5} rightrfloor = 20 )2. Multiples of 25: ( leftlfloor frac{100}{25} rightrfloor = 4 )3. Multiples of 125: ( leftlfloor frac{100}{125} rightrfloor = 0 )Summing these values gives the total number of factors of 5 in 100!:[ 20 + 4 + 0 = 24 ]Thus, the number of trailing zeros in 100! is (boxed{24}).]

🔑:4.35. Answer: 24. Among the numbers from 1 to 100, there are 20 numbers divisible by 5, and among the numbers divisible by 5, there are 4 numbers divisible by 25 (there are no numbers divisible by 125 among the given numbers). Therefore, the considered product is divisible by 5^{24} and not divisible by 5^{25}. It is also clear that it is divisible by 2^{24}.

🧠:[<think>Okay, so I need to solve the equation: (frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2). Hmm, let's start by understanding what this equation is asking. It has two rational expressions with denominators that are quadratic in (x), and the sum of these two fractions equals 2. My goal is to find all real numbers (x) that satisfy this equation.First, I should note that the denominators (x^2 + 4) and (x^2 + 5) are always positive because (x^2) is non-negative for all real (x). Therefore, there's no risk of division by zero here, since (x^2 + 4) is at least 4, and (x^2 + 5) is at least 5. So, the equation is defined for all real numbers (x). That's good to know; there are no restrictions on the domain other than (x) being real.Now, let's think about how to solve this equation. Since both terms on the left side are fractions with denominators involving (x^2), maybe I can make a substitution to simplify things. Let me let (y = x^2). Then, the equation becomes:[frac{4}{y + 4} + frac{5}{y + 5} = 2]This substitution reduces the equation to one involving (y), which might be easier to handle. Now, my task is to solve for (y), and then relate that back to (x) by taking square roots (since (y = x^2)).Let's work with the equation in terms of (y):[frac{4}{y + 4} + frac{5}{y + 5} = 2]To solve this, I'll first find a common denominator for the two fractions on the left side. The denominators are (y + 4) and (y + 5), so their common denominator is ((y + 4)(y + 5)). Let's combine the fractions:[frac{4(y + 5) + 5(y + 4)}{(y + 4)(y + 5)} = 2]Expanding the numerator:[frac{4y + 20 + 5y + 20}{(y + 4)(y + 5)} = 2]Combine like terms in the numerator:[frac{9y + 40}{(y + 4)(y + 5)} = 2]Now, multiply both sides of the equation by ((y + 4)(y + 5)) to eliminate the denominator:[9y + 40 = 2(y + 4)(y + 5)]Now, let's expand the right side. First, multiply out ((y + 4)(y + 5)):[(y + 4)(y + 5) = y^2 + 5y + 4y + 20 = y^2 + 9y + 20]Therefore, the equation becomes:[9y + 40 = 2(y^2 + 9y + 20)]Distribute the 2 on the right side:[9y + 40 = 2y^2 + 18y + 40]Now, let's bring all terms to one side to form a quadratic equation. Subtract (9y + 40) from both sides:[0 = 2y^2 + 18y + 40 - 9y - 40]Simplify:[0 = 2y^2 + 9y]Factor out a (y) from the right side:[0 = y(2y + 9)]So, the solutions for (y) are:1. (y = 0)2. (2y + 9 = 0 implies y = -frac{9}{2})But wait a second. Remember that (y = x^2), and (x^2) cannot be negative. So, (y = -frac{9}{2}) is not a valid solution because (x^2) can't be negative. Therefore, the only valid solution here is (y = 0).Now, since (y = x^2), we substitute back to find (x):[x^2 = 0 implies x = 0]Therefore, the solution is (x = 0). But wait, let's verify this solution in the original equation to make sure we didn't make any mistakes during our algebraic manipulations.Original equation:[frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2]Substitute (x = 0):Left side:[frac{4}{0 + 4} + frac{5}{0 + 5} = frac{4}{4} + frac{5}{5} = 1 + 1 = 2]Right side: 2So, yes, it checks out. The left side equals the right side when (x = 0). Therefore, (x = 0) is indeed a solution.But wait, let's double-check our algebra when solving for (y). Maybe there's an error in the steps that caused us to miss another solution? Let me retrace:Starting from:[9y + 40 = 2(y + 4)(y + 5)]Expanding the right side:[2(y^2 + 9y + 20) = 2y^2 + 18y + 40]Subtracting left side:[2y^2 + 18y + 40 - 9y - 40 = 2y^2 + 9y]Yes, that's correct. Then factoring:[2y^2 + 9y = 0 implies y(2y + 9) = 0]So, (y = 0) or (y = -9/2). But since (y = x^2), only (y = 0) is valid. So, there's only one solution here. So, no other solutions.But maybe I should consider if there are other solutions that we missed? For example, sometimes when we multiply both sides of an equation by something, we might introduce extraneous solutions, but in this case, since we multiplied by ((y + 4)(y + 5)), which is always positive (since (y = x^2 geq 0)), there's no danger of multiplying by zero or a negative value, so the step is valid. Hence, the equation reduces correctly, and only (y = 0) is the solution.Therefore, (x = 0) is the only solution. Let me just think if there's another approach that might lead to another solution, but perhaps not.Alternatively, perhaps we can graph both sides of the equation. The left side is the sum of two decreasing functions (since as (x) increases, (x^2) increases, so denominators increase, making each fraction decrease). Therefore, the left side is a decreasing function in (x^2), which would suggest that there might be only one solution where it intersects the constant function 2. Since we found (x = 0), which gives the maximum possible value for each term (since (x = 0) minimizes the denominators), so (frac{4}{4} + frac{5}{5} = 1 + 1 = 2), which is exactly the case. If (x) were any other value, the denominators would be larger, making each fraction smaller, hence their sum would be less than 2. Therefore, the only solution is (x = 0).Wait, but let's test (x = 1) just to confirm. If (x = 1), then:Left side:[frac{4}{1 + 4} + frac{5}{1 + 5} = frac{4}{5} + frac{5}{6} = 0.8 + 0.833... ≈ 1.633... < 2]So, indeed, it's less than 2. How about (x = -1)? Same as (x = 1) because (x^2) is same for both. So, left side is the same, 1.633...What if (x) approaches infinity? Then, both terms on the left side approach zero, so the sum approaches zero, which is much less than 2. Therefore, the function is decreasing as (|x|) increases, starting from 2 at (x = 0) and going to zero as (|x| to infty). Therefore, the equation is only satisfied at (x = 0). That seems to confirm our solution.But let me check another point closer to zero. Let's say (x = 0.5):Left side:[frac{4}{0.25 + 4} + frac{5}{0.25 + 5} = frac{4}{4.25} + frac{5}{5.25} ≈ 0.9412 + 0.9524 ≈ 1.8936 < 2]Still less than 2. So, even near zero, the sum is less than 2 except exactly at zero. Therefore, only (x = 0) works.Alternatively, if we think of this as a function (f(x) = frac{4}{x^2 + 4} + frac{5}{x^2 + 5}), then (f(0) = 2), and (f(x)) decreases as (|x|) increases. Therefore, the equation (f(x) = 2) has exactly one solution at (x = 0). So, this makes sense.Therefore, I think we can be confident that the only solution is (x = 0).But just to make sure, let me try to approach this problem without substitution. Let's go back to the original equation:[frac{4}{x^2 + 4} + frac{5}{x^2 + 5} = 2]Let me denote (A = x^2). Then the equation becomes:[frac{4}{A + 4} + frac{5}{A + 5} = 2]Which is the same substitution as before. So, even if we didn't use substitution, the steps would be similar. Let's see:Multiply both sides by ((A + 4)(A + 5)):[4(A + 5) + 5(A + 4) = 2(A + 4)(A + 5)]Which gives:[4A + 20 + 5A + 20 = 2(A^2 + 9A + 20)]Simplify left side:9A + 40 = 2A^2 + 18A + 40Subtract left side:0 = 2A^2 + 9AThus, (A(2A + 9) = 0), so (A = 0) or (A = -9/2). Again, since (A = x^2 geq 0), only (A = 0) is valid, leading to (x = 0).Therefore, same result. So, no mistakes there.Alternatively, maybe a different approach. Let's consider moving all terms to one side:[frac{4}{x^2 + 4} + frac{5}{x^2 + 5} - 2 = 0]Combine the terms into a single fraction. Let's compute:First, write 2 as (frac{2(x^2 + 4)(x^2 + 5)}{(x^2 + 4)(x^2 + 5)}). Wait, maybe that's too complicated, but let's try.Express each term with the common denominator ((x^2 + 4)(x^2 + 5)):[frac{4(x^2 + 5) + 5(x^2 + 4) - 2(x^2 + 4)(x^2 + 5)}{(x^2 + 4)(x^2 + 5)} = 0]The numerator must be zero. Let's compute the numerator:First, expand each term:4(x² + 5) = 4x² + 205(x² + 4) = 5x² + 20So, adding those two: 4x² + 20 + 5x² + 20 = 9x² + 40Now, subtract 2(x² + 4)(x² + 5). Let's expand that:First, compute (x² + 4)(x² + 5) = x⁴ + 5x² + 4x² + 20 = x⁴ + 9x² + 20Then, multiply by 2: 2x⁴ + 18x² + 40Therefore, the entire numerator is:9x² + 40 - (2x⁴ + 18x² + 40) = -2x⁴ -9x²So, the numerator simplifies to:-2x⁴ -9x² = 0Factor out -x²:- x²(2x² + 9) = 0Set equal to zero:- x²(2x² + 9) = 0Thus, solutions are:x² = 0 ⇒ x = 0or2x² + 9 = 0 ⇒ x² = -9/2But x² cannot be negative, so the only solution is x = 0.Therefore, same result again. So, regardless of the method, we arrive at x = 0 as the only solution. This gives me confidence that the solution is indeed correct.Therefore, the answer is x = 0.Final AnswerThe solution is (boxed{0}).</think>To solve the equation (frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2), we start by making a substitution (y = x^2). This transforms the equation into:[frac{4}{y + 4} + frac{5}{y + 5} = 2]Next, we find a common denominator ((y + 4)(y + 5)) and combine the fractions:[frac{4(y + 5) + 5(y + 4)}{(y + 4)(y + 5)} = 2]Expanding and simplifying the numerator:[frac{4y + 20 + 5y + 20}{(y + 4)(y + 5)} = 2 implies frac{9y + 40}{(y + 4)(y + 5)} = 2]Multiplying both sides by ((y + 4)(y + 5)):[9y + 40 = 2(y^2 + 9y + 20)]Expanding and simplifying:[9y + 40 = 2y^2 + 18y + 40 implies 0 = 2y^2 + 9y]Factoring out (y):[0 = y(2y + 9)]This gives solutions (y = 0) or (y = -frac{9}{2}). Since (y = x^2) must be non-negative, the only valid solution is (y = 0). Substituting back, we get (x^2 = 0 implies x = 0).Verifying (x = 0) in the original equation confirms it is a solution. Further analysis shows the left side of the equation is a decreasing function in (x^2), confirming (x = 0) is the only solution.Thus, the solution is (boxed{0}).,<think>Okay, let's see. I need to solve the equation (frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2). Hmm, this looks like a rational equation. I remember that to solve equations with variables in the denominator, one common method is to eliminate the denominators by multiplying both sides by the least common denominator. But in this case, the denominators are (x^2 + 4) and (x^2 + 5). Since they are different, maybe I need to combine the fractions first or find a common denominator?Let me write down the equation again to get a clear view:[frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2]Alright, so the denominators are similar but not the same. Let me denote (y = x^2). Since (x^2) is always non-negative, (y geq 0). Substituting (y) into the equation simplifies the equation to:[frac{4}{y + 4} + frac{5}{y + 5} = 2]That looks a bit simpler. Now, I can try to solve for (y) first and then relate back to (x). Let's work with this substitution.To combine the fractions, the common denominator would be ((y + 4)(y + 5)). Let's multiply each term by this common denominator to eliminate the fractions:[4(y + 5) + 5(y + 4) = 2(y + 4)(y + 5)]Let me expand each term step by step.First, the left side:- (4(y + 5) = 4y + 20)- (5(y + 4) = 5y + 20)Adding those together:(4y + 20 + 5y + 20 = 9y + 40)Now the right side:(2(y + 4)(y + 5))First, expand ((y + 4)(y + 5)):(y^2 + 5y + 4y + 20 = y^2 + 9y + 20)Multiply by 2:(2y^2 + 18y + 40)So now, the equation becomes:Left side: (9y + 40)Right side: (2y^2 + 18y + 40)Subtracting the left side from both sides to set the equation to zero:(2y^2 + 18y + 40 - 9y - 40 = 0)Simplify:(2y^2 + 9y = 0)Wait, let me check that again. Let's subtract left side from right side:(2y^2 + 18y + 40 - (9y + 40) = 0)So, (2y^2 + 18y + 40 - 9y - 40 = 0)Yes, that simplifies to:(2y^2 + 9y = 0)Wait, that seems too simple. Let me check if I made a mistake in expanding or combining terms.Original equation after substitution:(4/(y + 4) + 5/(y + 5) = 2)Multiply both sides by ((y + 4)(y + 5)):Left side: 4(y + 5) + 5(y + 4) = 4y + 20 + 5y + 20 = 9y + 40Right side: 2(y + 4)(y + 5) = 2(y^2 + 9y + 20) = 2y^2 + 18y + 40Then moving everything to one side:2y^2 + 18y + 40 - 9y - 40 = 0 ⇒ 2y^2 + 9y = 0Hmm, so 2y² + 9y = 0. Then factor out a y:y(2y + 9) = 0So either y = 0 or 2y + 9 = 0. Solving for y:Case 1: y = 0Case 2: 2y + 9 = 0 ⇒ y = -9/2But since y = x², and x² cannot be negative, we discard y = -9/2 as an extraneous solution. Therefore, the only valid solution is y = 0.Thus, x² = 0 ⇒ x = 0.Wait, but let me check if that's correct. Let me substitute x = 0 back into the original equation:Left side: 4/(0 + 4) + 5/(0 + 5) = 4/4 + 5/5 = 1 + 1 = 2Right side: 2So 2 = 2. That works. But is that the only solution? Let me check if I missed anything.When we multiplied both sides by (y + 4)(y + 5), we have to make sure that (y + 4)(y + 5) ≠ 0. Since y ≥ 0, y + 4 ≥ 4 and y + 5 ≥ 5, so denominators are never zero. Therefore, no extraneous solutions introduced there.But the quadratic equation gave us y = 0 and y = -9/2. But since y can't be negative, y = 0 is the only solution. Therefore, x = 0 is the only solution.But wait, is there a possibility that there are other solutions? Let me check.Suppose I didn't substitute y = x² and worked with the original equation. Let me try to solve it another way to verify.Original equation:[frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2]Let me denote (a = x^2 + 4) and (b = x^2 + 5). Then, since (b = a + 1), the equation becomes:[frac{4}{a} + frac{5}{a + 1} = 2]Multiply both sides by (a(a + 1)):Left side: 4(a + 1) + 5a = 4a + 4 + 5a = 9a + 4Right side: 2a(a + 1) = 2a² + 2aBring all terms to one side:2a² + 2a - 9a - 4 = 0 ⇒ 2a² -7a -4 = 0Wait, this is different from the previous equation. Wait, maybe I made a mistake here.Wait, let's check again. If a = x² +4, then x² +5 = a +1. Then the equation is 4/a +5/(a +1) =2.Multiply both sides by a(a +1):4(a +1) +5a = 2a(a +1)Left side: 4a +4 +5a = 9a +4Right side: 2a² + 2aThen, bringing all terms to left side: 9a +4 -2a² -2a =0 ⇒ -2a² +7a +4 =0Multiply both sides by -1: 2a² -7a -4 =0So quadratic equation: 2a² -7a -4=0Let me solve this quadratic equation.Using quadratic formula:a = [7 ± sqrt(49 + 32)] / 4 = [7 ± sqrt(81)] /4 = [7 ±9]/4So:a = (7 +9)/4 =16/4=4a= (7 -9)/4= -2/4= -1/2Since a = x² +4, which is x² +4 ≥4, so a must be ≥4. Therefore, a= -1/2 is invalid. So only a=4.Therefore, a=4 ⇒ x² +4=4 ⇒ x²=0 ⇒x=0.Same result as before. So that confirms that x=0 is the only solution.But wait, in the first substitution, we ended up with quadratic in y: 2y² +9y=0, leading to y=0. Here, by substitution with a, we ended up with a quadratic equation leading to a=4, which gives x=0. So both methods lead to the same solution. So that seems consistent.Therefore, x=0 is the solution. But let's just check once again to make sure.Plug x=0 into the original equation:4/(0 +4) +5/(0 +5)=4/4 +5/5=1 +1=2. Which matches the right side.But just to be thorough, let me check if there's a possibility of another solution. Suppose x is non-zero. Let's pick x=1:Left side: 4/(1 +4) +5/(1 +5)=4/5 +5/6≈0.8 +0.833≈1.633, which is less than 2.x=2: 4/(4 +4)=4/8=0.5; 5/(4 +5)=5/9≈0.555; total≈1.055 <2.x approaching infinity: as x becomes very large, 4/(x² +4)≈4/x² and 5/(x² +5)≈5/x². So total≈9/x², which approaches 0. So it won't reach 2.Negative x? x=-3: same as x=3, since x is squared. So same result. So even if x is negative, same as positive.What about x= something imaginary? Wait, the problem didn't specify if x is real or complex. But generally, unless stated otherwise, we assume real numbers. So x is real.Therefore, the only real solution is x=0.Wait, but just to check another value. Let's try x=0.5:Left side: 4/(0.25 +4)=4/4.25≈0.941; 5/(0.25 +5)=5/5.25≈0.952; sum≈1.893 <2.x= sqrt(something). Let me see. Suppose x²=1, we already checked x=1. x²=0.1: 4/(0.1 +4)=4/4.1≈0.9756; 5/(0.1 +5)=5/5.1≈0.9804; total≈1.956, still less than 2.x² approaching zero from the positive side: x²=0.0001:4/(0.0001 +4)=≈4/4.0001≈0.999975; 5/(0.0001 +5)=≈5/5.0001≈0.99998; sum≈1.99995, which is just below 2. So as x approaches zero, the sum approaches 2. But at x=0, it's exactly 2. Therefore, x=0 is the only point where it equals 2.Thus, seems like x=0 is indeed the only solution.Wait, but the first method gave 2y² +9y=0, which led to y=0 and y=-4.5. But when I did the substitution with a, it led to a quadratic equation with solution a=4 and a=-0.5. However, both approaches lead to the same x=0 solution. So seems like x=0 is the only solution.Therefore, after all this checking, I can conclude that the solution is x=0.Final AnswerThe solution is (boxed{0}).</think>To solve the equation (frac{4}{x^{2}+4} + frac{5}{x^{2}+5} = 2), we start by substituting (y = x^2). This transforms the equation into:[frac{4}{y + 4} + frac{5}{y + 5} = 2]Next, we multiply both sides by the common denominator ((y + 4)(y + 5)):[4(y + 5) + 5(y + 4) = 2(y + 4)(y + 5)]Expanding and simplifying both sides:- Left side: (4(y + 5) + 5(y + 4) = 4y + 20 + 5y + 20 = 9y + 40)- Right side: (2(y + 4)(y + 5) = 2(y^2 + 9y + 20) = 2y^2 + 18y + 40)Setting the equation to zero by subtracting the left side from the right side:[2y^2 + 18y + 40 - 9y - 40 = 0 implies 2y^2 + 9y = 0]Factoring out (y):[y(2y + 9) = 0]This gives solutions (y = 0) or (y = -frac{9}{2}). Since (y = x^2) must be non-negative, we discard (y = -frac{9}{2}). Thus, (y = 0) leads to (x^2 = 0), hence (x = 0).Verification by substituting (x = 0) back into the original equation confirms it is a solution:[frac{4}{0^2 + 4} + frac{5}{0^2 + 5} = frac{4}{4} + frac{5}{5} = 1 + 1 = 2]Thus, the only solution is (boxed{0}).]

🔑:## Solution.Domain: x in R.frac{2 x^{4}+9 x^{2}}{left(x^{2}+4right)left(x^{2}+5right)}=0 Leftrightarrow 2 x^{4}+9 x^{2}=0 Leftrightarrow x^{2}left(2 x^{2}+9right)=0,x^{2}=0, x_{1}=0 or 2 x^{2}+9=0, x_{2,3} in varnothing.Answer: x=0.