answer：Okay, so I have this problem: Find the number of pairs of positive integers (x, y) such that x squared minus y squared equals 45. Hmm, that sounds like a Diophantine equation. I remember that Diophantine equations are equations where we look for integer solutions. Alright, so the equation is x² - y² = 45. I think I can factor this equation because it's a difference of squares. Let me try that. So, x² - y² factors into (x + y)(x - y). That gives me (x + y)(x - y) = 45. Now, since x and y are positive integers, both (x + y) and (x - y) must also be positive integers. Also, since x > y (because x² - y² is positive), x - y is a positive integer. So, I can think of this as finding two positive integers, say a and b, such that a * b = 45, and a > b. Then, I can set a = x + y and b = x - y, and solve for x and y.So, first, I need to find all pairs of positive integers (a, b) where a * b = 45 and a > b. Let me list the positive divisors of 45. The divisors are 1, 3, 5, 9, 15, and 45. Now, I can pair these divisors such that their product is 45. Let me list all possible pairs (a, b):1. (45, 1)2. (15, 3)3. (9, 5)Wait, are there any more? Let me check. If I take 5 and 9, that's the same as 9 and 5, but since a > b, I already have that. Similarly, 3 and 15 is the same as 15 and 3, which I have. So, I think these are all the possible pairs where a > b.Now, for each pair (a, b), I can solve for x and y. Let's take the first pair: a = 45, b = 1.So, we have:x + y = 45x - y = 1I can solve these two equations simultaneously. If I add them together, I get:(x + y) + (x - y) = 45 + 12x = 46x = 23Then, substituting back into x + y = 45:23 + y = 45y = 45 - 23y = 22So, one solution is (23, 22).Next, let's take the second pair: a = 15, b = 3.So, we have:x + y = 15x - y = 3Adding these two equations:(x + y) + (x - y) = 15 + 32x = 18x = 9Substituting back into x + y = 15:9 + y = 15y = 15 - 9y = 6So, another solution is (9, 6).Now, the third pair: a = 9, b = 5.So, we have:x + y = 9x - y = 5Adding the equations:(x + y) + (x - y) = 9 + 52x = 14x = 7Substituting back into x + y = 9:7 + y = 9y = 9 - 7y = 2So, the third solution is (7, 2).Are there any more pairs? Let me think. If I consider a = 5 and b = 9, but since a must be greater than b, that's not necessary. Similarly, a = 3 and b = 15 would give negative y, which isn't allowed because y has to be positive. So, I think these three pairs are the only ones that work.Let me just verify each solution to make sure they satisfy the original equation.First, (23, 22):23² - 22² = (23 - 22)(23 + 22) = 1 * 45 = 45. Correct.Second, (9, 6):9² - 6² = (9 - 6)(9 + 6) = 3 * 15 = 45. Correct.Third, (7, 2):7² - 2² = (7 - 2)(7 + 2) = 5 * 9 = 45. Correct.So, all three solutions work. Therefore, there are three pairs of positive integers (x, y) that satisfy the equation x² - y² = 45.I think that's it. I don't think I missed any pairs because I considered all the factor pairs of 45 where a > b, and each gave a valid solution with positive integers x and y. So, the answer should be three pairs.

answer：Alright, so I have these two trigonometry problems to solve. Let me take them one at a time and think through each step carefully.**Problem 1:** Given that sinα = 5/13, find the values of cosα and tanα.Okay, I remember that in trigonometry, there's this fundamental identity called the Pythagorean identity, which says that sin²α + cos²α = 1. Since I know sinα, I can use this identity to find cosα.So, let's write that down:sin²α + cos²α = 1We know sinα is 5/13, so sin²α is (5/13)², which is 25/169. Plugging that into the equation:25/169 + cos²α = 1Now, to find cos²α, I subtract 25/169 from both sides:cos²α = 1 - 25/169Calculating 1 as 169/169 to have a common denominator:cos²α = 169/169 - 25/169 = 144/169So, cos²α is 144/169. To find cosα, I take the square root of both sides:cosα = ±√(144/169) = ±12/13Hmm, so cosα could be either 12/13 or -12/13. But how do I know which one it is? Well, that depends on the quadrant where angle α is located. Since the problem doesn't specify, I think I should consider both possibilities. So, cosα is either 12/13 or -12/13.Now, moving on to tanα. I remember that tanα is sinα divided by cosα. So:tanα = sinα / cosαWe have sinα = 5/13 and cosα = ±12/13. Plugging those in:tanα = (5/13) / (12/13) = 5/12Or, if cosα is -12/13:tanα = (5/13) / (-12/13) = -5/12So, tanα is either 5/12 or -5/12, depending on the value of cosα.Alright, that takes care of Problem 1.**Problem 2:** Given that tanα = 2, find the value of 1 / (2sinαcosα + cos²α).Hmm, this looks a bit more complex. Let me see. I need to simplify the expression in the denominator first. Maybe I can express everything in terms of tanα since I know tanα is 2.I remember that tanα = sinα / cosα, so sinα = tanα * cosα. Let's substitute that into the denominator:2sinαcosα + cos²α = 2(tanα * cosα)cosα + cos²αSimplify that:= 2tanα * cos²α + cos²αFactor out cos²α:= cos²α (2tanα + 1)So, the denominator becomes cos²α (2tanα + 1). Therefore, the entire expression is:1 / [cos²α (2tanα + 1)]Now, I need to express this in terms of tanα. I recall another identity: 1 + tan²α = sec²α, and since secα = 1/cosα, then cos²α = 1 / (1 + tan²α).Let me write that down:cos²α = 1 / (1 + tan²α)Given that tanα = 2, tan²α = 4. So,cos²α = 1 / (1 + 4) = 1/5Alright, so cos²α is 1/5. Now, let's substitute back into the expression:1 / [cos²α (2tanα + 1)] = 1 / [(1/5)(2*2 + 1)]Calculate 2*2 + 1:= 1 / [(1/5)(4 + 1)] = 1 / [(1/5)(5)] = 1 / 1 = 1Wait, that's interesting. The entire expression simplifies to 1. So, the value is 1.But let me double-check my steps to make sure I didn't make a mistake.1. I started by expressing sinα in terms of tanα and cosα.2. Substituted into the denominator and factored out cos²α.3. Used the identity 1 + tan²α = sec²α to express cos²α in terms of tanα.4. Plugged in tanα = 2 and simplified.Everything seems to check out. So, the value is indeed 1.**Final Answer**Problem 1: cosα = ±12/13 and tanα = ±5/12.Problem 2: The value is 1.So, putting it all together:Problem 1:cosα = boxed{pm dfrac{12}{13}} tanα = boxed{pm dfrac{5}{12}}Problem 2:boxed{1}

answer：First, I need to evaluate the expression (2 + (4 cdot 3 - 2) div 2 cdot 3 + 5).I'll start by solving the operations inside the parentheses. Calculating (4 cdot 3) gives me 12, and then subtracting 2 results in 10.Now, the expression becomes (2 + 10 div 2 cdot 3 + 5).Next, I'll perform the division and multiplication from left to right. Dividing 10 by 2 gives me 5, and then multiplying by 3 results in 15.The expression is now simplified to (2 + 15 + 5).Finally, I'll add the remaining numbers: (2 + 15) equals 17, and adding 5 gives me the final result of 22.

answer：Okay, so I have this problem about a sequence {a_n} where a₁ = -3, and the recurrence relation is a_{n+1} + a_n = 4n - 2 for n in natural numbers. There are two parts: first, I need to prove that {a_n - 2(n - 1)} is a geometric sequence and find the general formula for {a_n}. Second, I need to define T_{2n} as the sum of the first 2n terms of {a_n}, then define b_n = T_{2n} + 6n, and prove that the sum of the first n terms of {1/b_n} is less than 1/4.Starting with part (1). I need to show that {a_n - 2(n - 1)} is a geometric sequence. Hmm, geometric sequences have the property that each term is a constant multiple of the previous term. So, if I can show that (a_{n+1} - 2n) is a constant multiple of (a_n - 2(n - 1)), then I can prove it's geometric.Given the recurrence relation a_{n+1} + a_n = 4n - 2. Maybe I can rearrange this to express a_{n+1} in terms of a_n. Let's try that:a_{n+1} = 4n - 2 - a_n.Now, let's consider the expression a_{n+1} - 2n. Substituting the expression for a_{n+1}:a_{n+1} - 2n = (4n - 2 - a_n) - 2n = (4n - 2 - 2n) - a_n = (2n - 2) - a_n.Hmm, that's 2(n - 1) - a_n. Wait, let me see:Wait, 4n - 2 - a_n - 2n = (4n - 2n) - 2 - a_n = 2n - 2 - a_n.So, a_{n+1} - 2n = 2(n - 1) - a_n.But I want to relate this to a_n - 2(n - 1). Let me write that:Let’s denote c_n = a_n - 2(n - 1). Then, c_{n+1} = a_{n+1} - 2n.From above, a_{n+1} - 2n = 2(n - 1) - a_n.So, c_{n+1} = 2(n - 1) - a_n.But c_n = a_n - 2(n - 1), so a_n = c_n + 2(n - 1).Substituting back into c_{n+1}:c_{n+1} = 2(n - 1) - (c_n + 2(n - 1)) = 2(n - 1) - c_n - 2(n - 1) = -c_n.So, c_{n+1} = -c_n. That means each term is -1 times the previous term. So, {c_n} is a geometric sequence with common ratio -1.Now, to find the first term c₁:c₁ = a₁ - 2(1 - 1) = a₁ - 0 = a₁ = -3.So, the general term of the geometric sequence {c_n} is c_n = c₁ * (-1)^{n-1} = -3 * (-1)^{n-1}.Therefore, a_n = c_n + 2(n - 1) = -3 * (-1)^{n-1} + 2(n - 1).Simplify that:-3 * (-1)^{n-1} can be written as 3 * (-1)^n because (-1)^{n-1} = (-1)^n * (-1)^{-1} = (-1)^n * (-1) = -(-1)^n. So, -3 * (-1)^{n-1} = 3 * (-1)^n.Thus, a_n = 3*(-1)^n + 2(n - 1).So, that's the general formula for {a_n}.Moving on to part (2). I need to find T_{2n}, which is the sum of the first 2n terms of {a_n}. Then define b_n = T_{2n} + 6n, and show that the sum of the first n terms of {1/b_n} is less than 1/4.First, let's compute T_{2n} = sum_{k=1}^{2n} a_k.Given that a_k = 3*(-1)^k + 2(k - 1).So, T_{2n} = sum_{k=1}^{2n} [3*(-1)^k + 2(k - 1)] = 3*sum_{k=1}^{2n} (-1)^k + 2*sum_{k=1}^{2n} (k - 1).Let's compute each sum separately.First sum: sum_{k=1}^{2n} (-1)^k.This is an alternating series: -1 + 1 -1 + 1 - ... for 2n terms.Since it's 2n terms, which is even, the sum will be 0 because each pair (-1 +1) cancels out.So, sum_{k=1}^{2n} (-1)^k = 0.Second sum: sum_{k=1}^{2n} (k - 1) = sum_{m=0}^{2n -1} m = sum from m=0 to m=2n-1 of m.The sum of the first m integers starting from 0 is (m)(m + 1)/2. Here, m = 2n -1.So, sum_{m=0}^{2n -1} m = (2n -1)(2n)/2 = n(2n -1).Therefore, T_{2n} = 3*0 + 2*n(2n -1) = 2n(2n -1) = 4n² - 2n.Now, define b_n = T_{2n} + 6n = (4n² - 2n) + 6n = 4n² + 4n.So, b_n = 4n² + 4n.Now, we need to consider the sequence {1/b_n} = {1/(4n² + 4n)}.We can factor out 4n from the denominator: 1/(4n(n +1)).So, 1/b_n = 1/(4n(n +1)).We can write this as (1/4)(1/(n(n +1))).Recall that 1/(n(n +1)) can be expressed as 1/n - 1/(n +1). This is a telescoping series.So, 1/(n(n +1)) = 1/n - 1/(n +1).Therefore, 1/b_n = (1/4)(1/n - 1/(n +1)).Now, we need to find the sum of the first n terms of {1/b_n}, which is sum_{k=1}^{n} 1/b_k.Substituting the expression we have:sum_{k=1}^{n} 1/b_k = sum_{k=1}^{n} (1/4)(1/k - 1/(k +1)) = (1/4) sum_{k=1}^{n} (1/k - 1/(k +1)).This is a telescoping series. Let's write out the terms:(1/4)[(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n +1))].When we add these up, most terms cancel out:- The -1/2 cancels with +1/2,- The -1/3 cancels with +1/3,- And so on, until the -1/n cancels with +1/n.What remains is the first term of the first expression and the last term of the last expression:(1/4)[1 - 1/(n +1)].Simplify this:(1/4)[(n +1 -1)/(n +1)] = (1/4)(n/(n +1)) = n/(4(n +1)).Now, we need to show that this sum is less than 1/4.So, n/(4(n +1)) < 1/4.Multiply both sides by 4:n/(n +1) < 1.Which is true because n < n +1, so n/(n +1) is always less than 1 for positive integers n.Therefore, the sum of the first n terms of {1/b_n} is n/(4(n +1)) < 1/4.Hence, proved.

answer：Alright, so I have this problem with two quadratic equations, and I need to find the range of m such that at least one of them has real roots. Hmm, okay. Let me try to break this down step by step.First, I remember that for a quadratic equation ax² + bx + c = 0 to have real roots, the discriminant must be non-negative. The discriminant is given by Δ = b² - 4ac. If Δ is greater than or equal to zero, the equation has real roots. So, I need to check the discriminants of both equations and see when at least one of them is non-negative.Let me write down the two equations again to make sure I have them correctly:1. x² + 4mx + 4m² + 2m + 3 = 02. x² + (2m + 1)x + m² = 0Okay, let's start with the first equation. I'll compute its discriminant.For the first equation, a = 1, b = 4m, and c = 4m² + 2m + 3.So, Δ₁ = b² - 4ac = (4m)² - 4 * 1 * (4m² + 2m + 3)Calculating that:Δ₁ = 16m² - 4*(4m² + 2m + 3)= 16m² - 16m² - 8m - 12= -8m - 12Hmm, so Δ₁ = -8m - 12. For the first equation to have real roots, Δ₁ ≥ 0.So, -8m - 12 ≥ 0Let me solve for m:-8m - 12 ≥ 0Add 12 to both sides:-8m ≥ 12Divide both sides by -8, remembering to flip the inequality sign because I'm dividing by a negative number:m ≤ -12 / 8Simplify:m ≤ -3/2Okay, so the first equation has real roots when m is less than or equal to -3/2.Now, moving on to the second equation. Let's compute its discriminant.For the second equation, a = 1, b = (2m + 1), and c = m².So, Δ₂ = b² - 4ac = (2m + 1)² - 4 * 1 * m²Calculating that:Δ₂ = (4m² + 4m + 1) - 4m²= 4m² + 4m + 1 - 4m²= 4m + 1So, Δ₂ = 4m + 1. For the second equation to have real roots, Δ₂ ≥ 0.So, 4m + 1 ≥ 0Solving for m:4m + 1 ≥ 0Subtract 1 from both sides:4m ≥ -1Divide both sides by 4:m ≥ -1/4Alright, so the second equation has real roots when m is greater than or equal to -1/4.Now, the problem states that at least one of the equations has real roots. So, this means that either the first equation has real roots, or the second equation has real roots, or both.From above, we have:- First equation has real roots when m ≤ -3/2- Second equation has real roots when m ≥ -1/4So, combining these two conditions, the values of m for which at least one equation has real roots are m ≤ -3/2 or m ≥ -1/4.Wait, let me think about this again. If m is less than or equal to -3/2, then the first equation has real roots, regardless of the second equation. Similarly, if m is greater than or equal to -1/4, the second equation has real roots, regardless of the first equation. So, the union of these two intervals is m ≤ -3/2 or m ≥ -1/4.Looking at the options given:A: -3/2 < m < -1/4B: m ≤ -3/2 or m ≥ -1/4C: -1/4 < m < 1/2D: m ≤ -3/2 or m ≥ 1/2So, option B matches our conclusion. Therefore, the correct answer should be B.But wait, let me double-check my calculations to make sure I didn't make any mistakes.For the first equation:Δ₁ = (4m)² - 4*(4m² + 2m + 3)= 16m² - 16m² - 8m - 12= -8m - 12Yes, that's correct. So, setting Δ₁ ≥ 0:-8m - 12 ≥ 0-8m ≥ 12m ≤ -12/8m ≤ -3/2That's correct.For the second equation:Δ₂ = (2m + 1)² - 4m²= 4m² + 4m + 1 - 4m²= 4m + 1Yes, that's correct. So, setting Δ₂ ≥ 0:4m + 1 ≥ 04m ≥ -1m ≥ -1/4That's correct as well.So, combining these, m ≤ -3/2 or m ≥ -1/4, which is option B.I think that's solid. I don't see any mistakes in my reasoning. So, I'm confident that the answer is B.**Final Answer**boxed{B}

answer：Alright, so I've got this triangle problem here, and I need to solve two parts. Let me take it step by step.First, part (1) says: In triangle ABC, the sides opposite angles A, B, and C are a, b, and c respectively. It's given that sin(A + π/6) = 2 cos A. I need to find the value of angle A.Okay, so I remember some trigonometric identities. Maybe I can expand sin(A + π/6) using the sine addition formula. The formula is sin(x + y) = sin x cos y + cos x sin y. So, applying that here:sin(A + π/6) = sin A cos(π/6) + cos A sin(π/6).I know that cos(π/6) is √3/2 and sin(π/6) is 1/2. So substituting those values in:sin(A + π/6) = sin A*(√3/2) + cos A*(1/2).According to the problem, this equals 2 cos A. So:(√3/2) sin A + (1/2) cos A = 2 cos A.Hmm, let me write that equation down:(√3/2) sin A + (1/2) cos A = 2 cos A.I can subtract (1/2) cos A from both sides to get:(√3/2) sin A = (2 - 1/2) cos A.Simplify the right side: 2 - 1/2 is 3/2. So:(√3/2) sin A = (3/2) cos A.Now, I can multiply both sides by 2 to eliminate the denominators:√3 sin A = 3 cos A.Divide both sides by cos A to get tan A:√3 tan A = 3.So tan A = 3 / √3 = √3.Okay, tan A = √3. I remember that tan(π/3) = √3, so A must be π/3. But wait, since A is an angle in a triangle, it must be between 0 and π, so π/3 is the only solution. So, A = π/3 radians, which is 60 degrees.Alright, that seems straightforward. Let me just verify my steps:1. Expanded sin(A + π/6) correctly.2. Substituted the known sine and cosine values for π/6.3. Set the equation equal to 2 cos A as given.4. Subtracted (1/2) cos A from both sides.5. Simplified the equation to find tan A = √3.6. Concluded A = π/3.Everything seems to check out. So, part (1) is solved, A is π/3.Moving on to part (2). It says: If a = √3 and the height from BC is 2/3, find the value of b + c.Wait, the height from BC. Hmm. In triangle ABC, the height from BC would be the altitude drawn from vertex A to side BC. So, the height h from A to BC is 2/3.Given that, and side a is opposite angle A, which is BC. So, side a is BC, which is √3. The height h from A to BC is 2/3.I remember that the area of a triangle can be expressed in two ways: one using the base and height, and another using two sides and the sine of the included angle.First, let's write the area using base BC and height h:Area = (1/2) * BC * h = (1/2) * √3 * (2/3).Calculating that: (1/2) * √3 * (2/3) = (√3)/3.So, the area is √3 / 3.Alternatively, the area can also be expressed as (1/2) * b * c * sin A.We know angle A is π/3, so sin A = sin(π/3) = √3/2.So, Area = (1/2) * b * c * (√3/2) = (b * c * √3)/4.We already found the area is √3 / 3, so:(b * c * √3)/4 = √3 / 3.Let me solve for bc:Multiply both sides by 4: b * c * √3 = (4√3)/3.Divide both sides by √3: b * c = 4/3.So, bc = 4/3.Alright, so now we have bc = 4/3.We also know side a = √3, which is opposite angle A = π/3.I think I can use the Law of Cosines here. The Law of Cosines states that:a² = b² + c² - 2bc cos A.Plugging in the known values:(√3)² = b² + c² - 2*(4/3)*cos(π/3).Calculate (√3)²: that's 3.cos(π/3) is 1/2, so:3 = b² + c² - 2*(4/3)*(1/2).Simplify the term with 2*(4/3)*(1/2): 2 and 1/2 cancel out, so it's 4/3.So, 3 = b² + c² - 4/3.Let me add 4/3 to both sides:3 + 4/3 = b² + c².Convert 3 to 9/3: 9/3 + 4/3 = 13/3.So, b² + c² = 13/3.Now, I need to find b + c. Hmm, how can I relate b + c to b² + c²?I remember that (b + c)² = b² + 2bc + c².We already know b² + c² = 13/3 and bc = 4/3.So, (b + c)² = 13/3 + 2*(4/3) = 13/3 + 8/3 = 21/3 = 7.Therefore, (b + c)² = 7, so b + c = √7.Wait, but since b and c are lengths, they must be positive, so we take the positive square root.So, b + c = √7.Let me recap the steps to make sure I didn't make a mistake:1. Calculated the area using base BC and height h: √3 / 3.2. Expressed area using sides b, c, and angle A: (b c √3)/4.3. Equated the two expressions and solved for bc: bc = 4/3.4. Applied the Law of Cosines: a² = b² + c² - 2bc cos A.5. Plugged in the known values and solved for b² + c²: 13/3.6. Used the identity (b + c)² = b² + 2bc + c² to find (b + c)² = 7.7. Took the square root to find b + c = √7.Everything seems to add up. I don't see any errors in my calculations. So, part (2) is solved, and b + c is √7.**Final Answer**(1) The value of angle ( A ) is (boxed{dfrac{pi}{3}}).(2) The value of ( b + c ) is (boxed{sqrt{7}}).