answer：Okay, so I need to find the range of the function ( y = frac{x^2 + 5x + 6}{x+2} ). Hmm, let's see. I remember that the range of a function is all the possible output values (y-values) it can produce. To find the range, I might need to analyze the function and see what values y can take on.First, I notice that the function is a rational function because it's a polynomial divided by another polynomial. The numerator is a quadratic, and the denominator is linear. I think I should start by simplifying the function if possible.Let me try factoring the numerator. The numerator is ( x^2 + 5x + 6 ). To factor this, I need two numbers that multiply to 6 and add up to 5. Let's see, 2 and 3 work because 2 * 3 = 6 and 2 + 3 = 5. So, I can factor the numerator as ( (x + 2)(x + 3) ).Now, the function becomes:[ y = frac{(x + 2)(x + 3)}{x + 2} ]Oh, I see that ( (x + 2) ) is a common factor in both the numerator and the denominator. So, I can simplify this by canceling out ( (x + 2) ), but I have to remember that ( x ) cannot be -2 because that would make the denominator zero, which is undefined.After canceling, the function simplifies to:[ y = x + 3 ]But with the restriction that ( x neq -2 ).So, if ( x ) can be any real number except -2, what does that mean for ( y )? Well, ( y = x + 3 ) is a linear function, which normally has a range of all real numbers. However, since ( x ) cannot be -2, ( y ) cannot be ( (-2) + 3 = 1 ). Therefore, ( y ) can be any real number except 1.To express this in interval notation, all real numbers except 1 would be written as two intervals: from negative infinity to 1, and from 1 to positive infinity. So, the range is ( (-infty, 1) cup (1, infty) ).Wait, let me double-check. If I plug in values around -2, does ( y ) approach 1? Let's see, if ( x ) approaches -2 from the left, say ( x = -2.1 ), then ( y = -2.1 + 3 = 0.9 ). If ( x ) approaches -2 from the right, say ( x = -1.9 ), then ( y = -1.9 + 3 = 1.1 ). So, as ( x ) gets closer to -2, ( y ) approaches 1 but never actually reaches it because ( x ) can't be -2.Therefore, my conclusion seems correct. The range of the function is all real numbers except 1, which in interval notation is ( (-infty, 1) cup (1, infty) ).

answer：Alright, so I've got this inequality to prove: for any points A, B, C, D, E on the plane, the sum of the distances AB + CD + DE + EC is less than or equal to the sum AC + AD + AE + BC + BD + BE. Hmm, okay. Let me try to wrap my head around this.First off, I need to understand what this inequality is saying. It's comparing two different sums of distances between these five points. On the left side, we're adding up AB, CD, DE, and EC. On the right side, we're adding up AC, AD, AE, BC, BD, and BE. So, it's like we're comparing a specific combination of distances on the left to a more comprehensive set of distances on the right.I wonder if there's a geometric interpretation here. Maybe something to do with triangles or polygons? Let me think about how these points could be arranged. If I imagine points A, B, C, D, E on a plane, they could form various shapes, but since the inequality has to hold for any arrangement, it must be a general property.Maybe I can start by considering a simpler case where all the points lie on a straight line. That might make things easier to visualize. Let's say A, B, C, D, E are all on a line in that order. Then, the distances would just be the absolute differences between their positions.So, if I denote their positions on the line as a, b, c, d, e respectively, then AB would be |b - a|, CD would be |d - c|, DE would be |e - d|, and EC would be |c - e|. On the right side, AC is |c - a|, AD is |d - a|, AE is |e - a|, BC is |c - b|, BD is |d - b|, and BE is |e - b|.Wait, if they're on a straight line, maybe I can express some of these distances in terms of others. For example, EC would be |c - e|, which is the same as |e - c|, so that's DE + DC if E is between D and C. Hmm, but in my initial assumption, I said they're in the order A, B, C, D, E. So, E is after D, which is after C. So, EC would actually be |e - c|, which is equal to ED + DC. But in the left side, we have DE and EC, which would be DE + EC = DE + (ED + DC) = 2DE + DC. Wait, that doesn't seem right.Maybe I messed up the order. Let me clarify: if they are in the order A, B, C, D, E on the line, then EC would be from E to C, which would be ED + DC. But DE is from D to E, which is just the same as ED. So, DE + EC would be DE + (ED + DC) = 2DE + DC. Hmm, but on the left side, we have AB + CD + DE + EC, which would be AB + CD + DE + (DE + DC) = AB + CD + 2DE + DC.On the right side, we have AC + AD + AE + BC + BD + BE. Let's express these in terms of the positions:AC = |c - a|AD = |d - a|AE = |e - a|BC = |c - b|BD = |d - b|BE = |e - b|Since they are in order A, B, C, D, E, all these distances are just the differences between the positions:AC = c - aAD = d - aAE = e - aBC = c - bBD = d - bBE = e - bSo, the right side is (c - a) + (d - a) + (e - a) + (c - b) + (d - b) + (e - b). Let's compute that:(c - a) + (d - a) + (e - a) + (c - b) + (d - b) + (e - b) = 3c + 3d + 3e - 3a - 2bWait, that seems a bit messy. Maybe I should compute it step by step:AC + AD + AE = (c - a) + (d - a) + (e - a) = c + d + e - 3aBC + BD + BE = (c - b) + (d - b) + (e - b) = c + d + e - 3bSo, adding them together: (c + d + e - 3a) + (c + d + e - 3b) = 2c + 2d + 2e - 3a - 3bNow, the left side was AB + CD + DE + EC. Let's compute that:AB = b - aCD = d - cDE = e - dEC = c - e (but since E is after C, EC would be e - c, right? Wait, no. If E is after D, which is after C, then EC is from E to C, which is e - c.Wait, but earlier I thought EC was DE + DC, but if E is after D, then EC is ED + DC, which is (e - d) + (d - c) = e - c. So, EC is just e - c. So, DE + EC would be (e - d) + (e - c) = 2e - d - c.Wait, but that doesn't seem right because DE is e - d and EC is e - c, so adding them gives 2e - d - c. But CD is d - c, so the left side is AB + CD + DE + EC = (b - a) + (d - c) + (e - d) + (e - c) = b - a + d - c + e - d + e - c = b - a + 2e - 2c.So, left side is b - a + 2e - 2c, and the right side is 2c + 2d + 2e - 3a - 3b.Wait, that doesn't seem to make sense because the left side is linear in b and e, while the right side has more terms. Maybe I made a mistake in computing the right side.Let me try again. The right side is AC + AD + AE + BC + BD + BE.AC = c - aAD = d - aAE = e - aBC = c - bBD = d - bBE = e - bSo, adding them up:(c - a) + (d - a) + (e - a) + (c - b) + (d - b) + (e - b) = c + d + e - 3a + c + d + e - 3b = 2c + 2d + 2e - 3a - 3bOkay, that's correct.Now, the left side is AB + CD + DE + EC = (b - a) + (d - c) + (e - d) + (e - c) = b - a + d - c + e - d + e - c = b - a + 2e - 2cSo, we have:Left side: b - a + 2e - 2cRight side: 2c + 2d + 2e - 3a - 3bNow, to compare these two, let's see if the left side is less than or equal to the right side.So, we need to check if:b - a + 2e - 2c ≤ 2c + 2d + 2e - 3a - 3bLet's move all terms to one side:b - a + 2e - 2c - 2c - 2d - 2e + 3a + 3b ≤ 0Simplify:(b + 3b) + (-a + 3a) + (2e - 2e) + (-2c - 2c) + (-2d) ≤ 0Which is:4b + 2a - 4c - 2d ≤ 0Divide both sides by 2:2b + a - 2c - d ≤ 0Hmm, so for this to hold, we need 2b + a ≤ 2c + dBut in our initial assumption, the points are in the order A, B, C, D, E on the line. So, a < b < c < d < e.Therefore, 2b + a ≤ 2c + d?Let's see: Since a < b < c < d, we can say that a < c and b < c, and d > c.So, 2b + a: since b < c and a < c, 2b + a < 2c + c = 3cOn the other side, 2c + d: since d > c, 2c + d > 2c + c = 3cSo, 2b + a < 3c < 2c + dTherefore, 2b + a < 2c + d, which implies 2b + a - 2c - d < 0So, the inequality holds in this case.Okay, so in the case where all points are on a straight line in the order A, B, C, D, E, the inequality holds.But the problem says "for any points A, B, C, D, E on the plane," so it's not limited to colinear points. I need to think about how to generalize this.Maybe I can use some geometric inequalities or properties. Perhaps the triangle inequality? Or maybe something like the Cauchy-Schwarz inequality?Wait, another approach could be to consider vectors or coordinates. Maybe assign coordinates to the points and express the distances in terms of coordinates, then try to manipulate the inequality.But that might get complicated with five points. Maybe there's a smarter way.Alternatively, I recall that in some inequalities involving multiple points, considering projections onto lines can be useful. Maybe I can use some kind of integral over all possible directions or something like that.Wait, I think I've heard of something called the "integral geometry" approach, where you integrate projections over all angles to derive inequalities. Maybe that's applicable here.Let me try to recall. If I project all points onto a line at some angle, then the sum of projections can be related to the original distances. Since the inequality has to hold for any configuration, integrating over all possible angles might help.So, suppose I fix a point O on the plane and consider rotating a line around O. For each angle φ, I can project all points onto this line and compute the distances in the projection. Then, integrating over all φ might give me an inequality that holds for the original distances.Let's formalize this a bit. Let l_φ be a line through O at angle φ. For any point X, let X_φ be its projection onto l_φ. Then, for any segment XY, the projected distance X_φY_φ is equal to XY times |cos(θ - φ)|, where θ is the angle of XY with respect to l_0 (some fixed line).So, if I integrate both sides of the inequality over φ from 0 to π, I can relate the sums of the original distances to the sums of their projections.But wait, how does that help? Let me think.If I integrate the left side and the right side over all φ, I can use the fact that the integral of |cos(φ - θ)| over φ from 0 to π is a constant, which might allow me to compare the sums.Let me try to write this out.For the left side: AB + CD + DE + ECEach term XY contributes XY * |cos(φ - θ_XY)| to the integral, where θ_XY is the angle of segment XY.Similarly, for the right side: AC + AD + AE + BC + BD + BEEach term contributes AC * |cos(φ - θ_AC)|, etc.So, integrating both sides over φ from 0 to π:∫₀^π (AB + CD + DE + EC) dφ = ∫₀^π (AB |cos(φ - θ_AB)| + CD |cos(φ - θ_CD)| + DE |cos(φ - θ_DE)| + EC |cos(φ - θ_EC)|) dφSimilarly for the right side:∫₀^π (AC + AD + AE + BC + BD + BE) dφ = ∫₀^π (AC |cos(φ - θ_AC)| + AD |cos(φ - θ_AD)| + AE |cos(φ - θ_AE)| + BC |cos(φ - θ_BC)| + BD |cos(φ - θ_BD)| + BE |cos(φ - θ_BE)|) dφNow, for each term, the integral of |cos(φ - θ)| over φ from 0 to π is 2, because:∫₀^π |cos(φ - θ)| dφ = ∫_{-θ}^{π - θ} |cos ψ| dψ = ∫₀^π |cos ψ| dψ = 2Since |cos ψ| has a period of π, the integral over any interval of length π is the same.Therefore, each distance term on both sides, when integrated over φ, contributes twice its length.So, the integral of the left side becomes 2(AB + CD + DE + EC), and the integral of the right side becomes 2(AC + AD + AE + BC + BD + BE).But from the original inequality, we have AB + CD + DE + EC ≤ AC + AD + AE + BC + BD + BE.Multiplying both sides by 2, we get 2(AB + CD + DE + EC) ≤ 2(AC + AD + AE + BC + BD + BE), which is exactly what the integrals give us.Therefore, since the integral of the left side is less than or equal to the integral of the right side, and the integral over all projections preserves the inequality, the original inequality must hold.Wait, does that make sense? I think I might be missing a step here. Let me double-check.We have that for each φ, the projected distances satisfy the inequality:AB_φ + CD_φ + DE_φ + EC_φ ≤ AC_φ + AD_φ + AE_φ + BC_φ + BD_φ + BE_φBecause for each projection, the inequality holds in the projected space, which is one-dimensional. In one dimension, we've already shown that the inequality holds when points are colinear.Therefore, integrating both sides over φ from 0 to π preserves the inequality, leading to:∫₀^π (AB_φ + CD_φ + DE_φ + EC_φ) dφ ≤ ∫₀^π (AC_φ + AD_φ + AE_φ + BC_φ + BD_φ + BE_φ) dφBut as we saw earlier, each integral of the projected distances equals twice the original distance. Therefore:2(AB + CD + DE + EC) ≤ 2(AC + AD + AE + BC + BD + BE)Dividing both sides by 2:AB + CD + DE + EC ≤ AC + AD + AE + BC + BD + BEWhich is exactly what we wanted to prove.So, by considering projections onto all possible lines through a fixed point and integrating, we've leveraged the fact that the inequality holds in one dimension to conclude it holds in two dimensions.I think that makes sense. It's a bit abstract, but using integral geometry seems like a powerful tool here. I should probably look up more about this technique to solidify my understanding, but for now, this approach seems to work.Another way to think about it is that since the inequality holds for any projection, and projections can be seen as averaging over all directions, the inequality must hold in the original plane as well.Yeah, I think that's the gist of it. So, by reducing the problem to one dimension and then integrating back up, we've proven the inequality for any configuration of points on the plane.

answer：Okay, so I have this problem about finding the greatest common divisor (gcd) of two numbers, x and y, where x is defined as 2^m - 1 and y is defined as 2^n + 1. The problem is divided into two parts: part (a) asks me to prove that if m/d is odd, then gcd(x, y) is 1, and part (b) asks me to find gcd(x, y) if m/d is even. Here, d is the gcd of m and n.First, I need to understand the problem. So, m and n are natural numbers, and d is their greatest common divisor. That means d is the largest number that divides both m and n. So, m can be written as d times some integer k, and n can be written as d times some integer l. So, m = d*k and n = d*l, where k and l are integers.Given that, x is 2^m - 1, which is 2^(d*k) - 1, and y is 2^n + 1, which is 2^(d*l) + 1. So, x = (2^d)^k - 1 and y = (2^d)^l + 1. Let me denote a = 2^d, so then x = a^k - 1 and y = a^l + 1.Now, for part (a), we are told that m/d is odd, which means k is odd. We need to prove that gcd(x, y) = 1.I remember that when dealing with gcds of numbers of the form a^k - 1 and a^l + 1, there are some properties we can use. One approach is to use the Euclidean algorithm, which involves repeatedly applying the gcd function with remainders.Let me recall that gcd(a^k - 1, a^l + 1) can sometimes be simplified by considering the exponents. Since a = 2^d, which is a power of 2, maybe there are some specific properties we can exploit here.I also remember that for numbers of the form a^k - 1 and a^l - 1, the gcd is a^gcd(k, l) - 1. But here, one of them is a^l + 1 instead of a^l - 1, so that might complicate things.Wait, maybe I can relate a^l + 1 to a^2l - 1. Because a^l + 1 is a factor of a^2l - 1, since (a^l - 1)(a^l + 1) = a^2l - 1. So, perhaps I can use that to find the gcd.So, let's consider gcd(x, y) = gcd(a^k - 1, a^l + 1). Since a^l + 1 divides a^2l - 1, we can say that gcd(a^k - 1, a^l + 1) divides gcd(a^k - 1, a^2l - 1). And as I mentioned earlier, gcd(a^k - 1, a^2l - 1) is equal to a^gcd(k, 2l) - 1.So, if I can find gcd(k, 2l), then I can find this gcd. Since k is odd (given in part (a)), and 2l is even, the gcd(k, 2l) would be gcd(k, l). Because k is odd, it doesn't share any factor of 2 with 2l, so it's just gcd(k, l).Therefore, gcd(a^k - 1, a^2l - 1) = a^gcd(k, l) - 1. So, now, since gcd(a^k - 1, a^l + 1) divides a^gcd(k, l) - 1, I need to see if this is equal to 1.But wait, a^gcd(k, l) - 1 is equal to (2^d)^gcd(k, l) - 1. So, if I can show that this is 1, then the gcd is 1. For that to happen, (2^d)^gcd(k, l) must be 2, which would mean that d*gcd(k, l) = 1. But d is a natural number, so d must be 1, and gcd(k, l) must be 1.But wait, is that necessarily the case? Or is there another way to see that the gcd is 1?Alternatively, maybe I can consider that since a = 2^d, and a ≡ 1 mod (a - 1). So, a^k ≡ 1 mod (a - 1), which means a^k - 1 ≡ 0 mod (a - 1). Similarly, a^l + 1 ≡ 1 + 1 = 2 mod (a - 1). So, the gcd of x and y must divide 2.But since a = 2^d, a - 1 = 2^d - 1. So, if d is 1, then a - 1 = 1, which doesn't help. But if d > 1, then a - 1 is at least 3, which is odd. So, 2 is not a multiple of a - 1, so the gcd must be 1.Wait, that seems a bit confusing. Let me try to structure this more clearly.Given that x = a^k - 1 and y = a^l + 1, where a = 2^d.We can consider that any common divisor of x and y must divide both x and y. So, suppose p is a prime that divides both x and y. Then, p divides x = a^k - 1 and p divides y = a^l + 1.Therefore, a^k ≡ 1 mod p and a^l ≡ -1 mod p.From the first congruence, the order of a modulo p divides k. From the second congruence, the order of a modulo p divides 2l because (a^l)^2 ≡ 1 mod p.So, the order of a modulo p divides both k and 2l. Since k is odd, the order must divide gcd(k, 2l). But since k is odd, gcd(k, 2l) = gcd(k, l). So, the order of a modulo p divides gcd(k, l).But from the first congruence, a^k ≡ 1 mod p, so the order divides k. From the second congruence, a^l ≡ -1 mod p, so squaring both sides, a^{2l} ≡ 1 mod p, so the order divides 2l.But since k is odd, and the order divides both k and 2l, it must divide gcd(k, 2l) = gcd(k, l). So, the order of a modulo p divides gcd(k, l).But also, from a^l ≡ -1 mod p, we can say that the order of a modulo p does not divide l, because a^l ≡ -1 ≠ 1 mod p. Therefore, the order must be exactly 2 times something. Wait, but if the order is d, then d divides 2l but does not divide l, so d must be even.But the order also divides k, which is odd. So, if the order is even, it cannot divide an odd number k. Therefore, the only possibility is that the order is 1. But if the order is 1, then a ≡ 1 mod p, which would mean that a^k ≡ 1 mod p and a^l ≡ 1 mod p. But y = a^l + 1 ≡ 1 + 1 = 2 mod p, so p divides 2. Therefore, p must be 2.But a = 2^d, so a is even, so a^k - 1 is odd (since 2^k is even, minus 1 is odd), and a^l + 1 is also odd (since 2^l is even, plus 1 is odd). Therefore, 2 cannot divide either x or y, so p cannot be 2. Therefore, there is no such prime p, which means that gcd(x, y) = 1.Okay, that makes sense. So, in part (a), if m/d is odd, then gcd(x, y) is 1.Now, moving on to part (b), where m/d is even. So, k is even. We need to find gcd(x, y).Using similar reasoning as before, let's consider that x = a^k - 1 and y = a^l + 1, where a = 2^d and k is even.Again, suppose p is a prime that divides both x and y. Then, a^k ≡ 1 mod p and a^l ≡ -1 mod p.From the first congruence, the order of a modulo p divides k, which is even. From the second congruence, the order divides 2l, as before.Since k is even, the order could potentially be 2. Let's see.If the order is 2, then a^2 ≡ 1 mod p, so a ≡ ±1 mod p. But a = 2^d, so 2^d ≡ 1 or -1 mod p.If 2^d ≡ 1 mod p, then a^k = (2^d)^k ≡ 1^k ≡ 1 mod p, which is consistent with x ≡ 0 mod p. Also, a^l = (2^d)^l ≡ 1^l ≡ 1 mod p, so y = a^l + 1 ≡ 1 + 1 = 2 mod p. Therefore, p divides 2, so p = 2. But as before, x and y are both odd, so p cannot be 2.If 2^d ≡ -1 mod p, then a = 2^d ≡ -1 mod p. Then, a^k = (-1)^k. Since k is even, a^k = 1 mod p, which is consistent with x ≡ 0 mod p. Also, a^l = (-1)^l. If l is odd, then a^l ≡ -1 mod p, so y = a^l + 1 ≡ -1 + 1 = 0 mod p. If l is even, then a^l ≡ 1 mod p, so y ≡ 1 + 1 = 2 mod p, which again would imply p = 2, which is not possible.Therefore, if l is odd, then p divides both x and y, and p must satisfy 2^d ≡ -1 mod p. So, p divides 2^{2d} - 1, because (2^d)^2 - 1 = 2^{2d} - 1.But wait, let's see. If 2^d ≡ -1 mod p, then 2^{2d} ≡ 1 mod p, so the order of 2 modulo p divides 2d. Also, since 2^d ≡ -1 mod p, the order does not divide d, so the order must be exactly 2d. Therefore, p must be a prime divisor of 2^{2d} - 1 but not of 2^d - 1.But 2^{2d} - 1 factors as (2^d - 1)(2^d + 1). So, p divides 2^d + 1. Therefore, any common prime divisor p of x and y must divide 2^d + 1.Therefore, the gcd(x, y) must divide 2^d + 1.Now, we need to check if 2^d + 1 actually divides both x and y.Let's see. Since x = 2^{d*k} - 1 and y = 2^{d*l} + 1.If 2^d ≡ -1 mod (2^d + 1), then 2^{d*k} ≡ (-1)^k mod (2^d + 1). Since k is even, (-1)^k = 1, so 2^{d*k} - 1 ≡ 1 - 1 = 0 mod (2^d + 1). Therefore, 2^d + 1 divides x.Similarly, 2^{d*l} ≡ (-1)^l mod (2^d + 1). If l is odd, then (-1)^l = -1, so 2^{d*l} + 1 ≡ -1 + 1 = 0 mod (2^d + 1). Therefore, if l is odd, 2^d + 1 divides y.But what if l is even? Then, (-1)^l = 1, so 2^{d*l} + 1 ≡ 1 + 1 = 2 mod (2^d + 1). Since 2^d + 1 is odd (as 2^d is even), 2 is not congruent to 0 mod (2^d + 1), so 2^d + 1 does not divide y in that case.Wait, but in part (b), we are only given that m/d is even, which is k is even. We don't have any condition on l, which is n/d. So, l could be either odd or even.But in the problem statement, part (b) just says "if m/d is even, find gcd(x, y)". So, perhaps we need to consider that regardless of l, the gcd is 2^d + 1.Wait, but if l is even, then y = 2^{d*l} + 1 = (2^{d*l/2})^2 + 1, which is similar to a sum of squares. But I don't know if that helps.Alternatively, maybe we can use the fact that gcd(x, y) divides 2^d + 1, and also, if l is odd, then 2^d + 1 divides y, so gcd(x, y) = 2^d + 1. If l is even, then y = 2^{d*l} + 1, and 2^{d*l} ≡ 1 mod (2^d + 1), so y ≡ 1 + 1 = 2 mod (2^d + 1). Therefore, gcd(x, y) would be gcd(2^d + 1, 2). Since 2^d + 1 is odd, gcd is 1.Wait, but that contradicts the earlier conclusion. So, perhaps the gcd is 2^d + 1 only if l is odd, and 1 if l is even.But the problem doesn't specify any condition on l, only on k. So, perhaps the answer depends on l.Wait, but in the problem, d is gcd(m, n), so l = n/d. Since m/d = k is even, and d is gcd(m, n), l must be co-prime with k. Because if k and l had a common divisor, then d wouldn't be the greatest common divisor of m and n.Wait, let me think. Since d = gcd(m, n), m = d*k, n = d*l, and gcd(k, l) = 1. Because if k and l had a common divisor, say t > 1, then d*t would divide both m and n, contradicting that d is the greatest common divisor.Therefore, gcd(k, l) = 1. So, in part (b), since k is even and gcd(k, l) = 1, l must be odd. Because if l were even, then k and l would both be even, contradicting gcd(k, l) = 1.Therefore, l must be odd. So, in part (b), l is necessarily odd because k is even and gcd(k, l) = 1.Therefore, y = 2^{d*l} + 1, and since l is odd, 2^{d*l} ≡ -1 mod (2^d + 1), so y ≡ -1 + 1 = 0 mod (2^d + 1). Therefore, 2^d + 1 divides y.Also, as before, 2^d + 1 divides x because k is even, so 2^{d*k} - 1 ≡ 1 - 1 = 0 mod (2^d + 1).Therefore, gcd(x, y) is at least 2^d + 1. But we also saw earlier that any common divisor must divide 2^d + 1, so gcd(x, y) = 2^d + 1.Therefore, in part (b), the gcd is 2^d + 1.So, to summarize:(a) If m/d is odd, then gcd(x, y) = 1.(b) If m/d is even, then gcd(x, y) = 2^d + 1.I think that makes sense. Let me just double-check.In part (a), since k is odd, and gcd(k, l) = 1, l could be either odd or even, but regardless, the reasoning showed that the gcd must be 1 because any common prime divisor would have to be 2, which isn't possible.In part (b), since k is even and gcd(k, l) = 1, l must be odd, so y = 2^{d*l} + 1 is divisible by 2^d + 1, and x is also divisible by 2^d + 1, so their gcd is 2^d + 1.Yes, that seems correct.

answer：Okay, so I have this sequence problem here. Let me read it again to make sure I understand what's being asked. In the sequence {a_n}, we have a₁ = 2 and a₁₇ = 66. The formula for the general term is a linear function of n. There are three parts to this problem:1. Find the general term formula for the sequence {a_n}.2. Calculate the value of a₂₀.3. Discuss whether 398 is an element of the sequence. Provide a reason.Alright, let's tackle each part step by step.Starting with part (1): Finding the general term formula. The problem says it's a linear function of n, which means the sequence is arithmetic. In an arithmetic sequence, each term increases by a constant difference, called the common difference, d.Given that a₁ = 2 and a₁₇ = 66, I can use these two terms to find the common difference, d. In an arithmetic sequence, the nth term is given by:a_n = a₁ + (n - 1)dSo, for a₁₇, which is the 17th term, n = 17. Plugging into the formula:a₁₇ = a₁ + (17 - 1)d66 = 2 + 16dNow, I can solve for d:66 - 2 = 16d64 = 16dd = 64 / 16d = 4Okay, so the common difference is 4. Now, plugging this back into the general formula:a_n = a₁ + (n - 1)da_n = 2 + (n - 1) * 4Let me simplify that:a_n = 2 + 4(n - 1)a_n = 2 + 4n - 4a_n = 4n - 2Wait, let me double-check that. If I expand 4(n - 1), that's 4n - 4. Then add 2: 4n - 4 + 2 = 4n - 2. Yeah, that seems right.So, the general term formula is a_n = 4n - 2.Moving on to part (2): Calculate the value of a₂₀.Using the formula I just found:a_n = 4n - 2So, for n = 20:a₂₀ = 4 * 20 - 2a₂₀ = 80 - 2a₂₀ = 78That seems straightforward. Let me just verify by calculating a few terms to see if the pattern holds.Given a₁ = 2, a₂ should be 6, a₃ should be 10, and so on. Each time, adding 4. So, a₁₇ should be 2 + 16*4 = 2 + 64 = 66, which matches the given information. So, the formula seems correct.Therefore, a₂₀ = 78.Now, part (3): Discuss whether 398 is an element of the sequence. Provide a reason.To determine if 398 is in the sequence, I need to see if there exists an integer n such that a_n = 398.Using the general term formula:a_n = 4n - 2Set this equal to 398:4n - 2 = 398Solve for n:4n = 398 + 24n = 400n = 400 / 4n = 100So, n = 100. Since n is a positive integer, 398 is indeed an element of the sequence. Specifically, it is the 100th term.Let me just make sure that n = 100 is valid. Since the sequence starts at n = 1 and increases by 1 each time, n = 100 is a valid term number. Therefore, 398 is part of the sequence.Wait, just to be thorough, let me plug n = 100 back into the formula to confirm:a₁₀₀ = 4*100 - 2 = 400 - 2 = 398Yep, that checks out. So, 398 is definitely in the sequence.So, summarizing:1. The general term is a_n = 4n - 2.2. a₂₀ = 78.3. 398 is an element of the sequence because when n = 100, a₁₀₀ = 398.I think that covers all parts of the problem. I don't see any mistakes in my calculations, and the logic seems sound. Each step follows from the previous one, and I've verified the results by plugging them back into the original formulas. So, I feel confident about these answers.**Final Answer**(1) The general term formula is boxed{4n - 2}.(2) The value of (a_{20}) is boxed{78}.(3) Yes, 398 is an element of the sequence because when (n = 100), (a_{100} = 398).

answer：Okay, so I have this problem about an isosceles triangle, triangle ABC, where the base angles A and B are equal. There's an altitude drawn from vertex C to the base, which divides angle C into two parts, C₁ and C₂. Additionally, angle C forms an external angle θ at vertex C, and θ is given as 30 degrees. I need to find the relationship between θ, C₁, and C₂.First, let me try to visualize the triangle. Since it's an isosceles triangle with base angles A and B equal, that means sides AC and BC are equal in length. So, vertex C is the apex, and AB is the base. The altitude from C to the base AB will split AB into two equal parts because it's an isosceles triangle. This altitude also splits angle C into two angles, C₁ and C₂.Now, the external angle θ at vertex C is given as 30 degrees. I remember that an external angle of a triangle is equal to the sum of the two non-adjacent internal angles. In this case, the external angle θ at vertex C should be equal to the sum of angles A and B. Since angles A and B are equal, let's denote each as α. So, θ = A + B = α + α = 2α. Given that θ is 30 degrees, we can set up the equation 2α = 30 degrees, which means α = 15 degrees. So, each of the base angles A and B is 15 degrees.Next, let's consider the altitude from vertex C to the base AB. This altitude creates two right triangles, each with a right angle at the point where the altitude meets the base. Let's denote the foot of the altitude as point D on AB. So, triangles ADC and BDC are both right-angled triangles.In triangle ADC, angle at D is 90 degrees, angle at A is 15 degrees, and angle at C is C₁. Similarly, in triangle BDC, angle at D is 90 degrees, angle at B is 15 degrees, and angle at C is C₂. Since the sum of angles in a triangle is 180 degrees, we can write the following equations for each right triangle:For triangle ADC:α + C₁ + 90° = 180°15° + C₁ + 90° = 180°C₁ = 180° - 15° - 90°C₁ = 75°Similarly, for triangle BDC:α + C₂ + 90° = 180°15° + C₂ + 90° = 180°C₂ = 180° - 15° - 90°C₂ = 75°So, both C₁ and C₂ are 75 degrees. Therefore, C₁ equals C₂, and each is 75 degrees.Let me double-check my reasoning. The external angle θ at vertex C is 30 degrees, which equals the sum of the two base angles A and B. Since A and B are equal, each is 15 degrees. The altitude splits angle C into two equal parts because the triangle is isosceles, so each part must be equal. Calculating each part gives 75 degrees, which makes sense because 15 degrees plus 75 degrees is 90 degrees, fitting the right triangle.I think this makes sense. So, the relationship is that both C₁ and C₂ are equal to 75 degrees each.

answer：Alright, so I've got this problem where a student found that five different trigonometric expressions all equal the same constant. I need to figure out what that constant is and then generalize it into a trigonometric identity. Let me take it step by step.First, I'll look at the five expressions given:1. ( sin^{2}13^{circ} + cos^{2}17^{circ} - sin13^{circ}cos17^{circ} )2. ( sin^{2}15^{circ} + cos^{2}15^{circ} - sin15^{circ}cos15^{circ} )3. ( sin^{2}18^{circ} + cos^{2}12^{circ} - sin18^{circ}cos12^{circ} )4. ( sin^{2}(-18^{circ}) + cos^{2}48^{circ} - sin(-18^{circ})cos48^{circ} )5. ( sin^{2}(-25^{circ}) + cos^{2}55^{circ} - sin(-25^{circ})cos55^{circ} )The first part asks me to choose one expression and calculate the constant. The second part is to generalize this into a trigonometric identity and prove it.Looking at these expressions, they all have a similar structure: ( sin^{2}A + cos^{2}B - sin A cos B ). The angles A and B in each case seem to add up to 30 degrees or something related. Let me check:1. 13° and 17°: 13 + 17 = 30°2. 15° and 15°: 15 + 15 = 30°3. 18° and 12°: 18 + 12 = 30°4. -18° and 48°: |-18| + 48 = 66°, which is not 30°, but wait, maybe it's related differently. Let me think.5. -25° and 55°: |-25| + 55 = 80°, which is also not 30°, but perhaps there's another relationship.Wait, maybe the angles are related in a different way. For expression 4, -18° and 48°, if I take 48° - (-18°) = 66°, which doesn't seem helpful. Alternatively, maybe 48° is 30° + 18°, and -18° is just negative. Similarly, for expression 5, 55° is 30° + 25°, and -25° is negative.So perhaps in each case, the angles are related such that B = 30° - A or something like that. Let me test this:For expression 1: A = 13°, B = 17°. 30° - 13° = 17°, so yes, B = 30° - A.Expression 2: A = 15°, B = 15°. 30° - 15° = 15°, so same as above.Expression 3: A = 18°, B = 12°. 30° - 18° = 12°, correct.Expression 4: A = -18°, B = 48°. 30° - (-18°) = 48°, correct.Expression 5: A = -25°, B = 55°. 30° - (-25°) = 55°, correct.So in all cases, B = 30° - A. That seems to be the pattern. So the general form is ( sin^{2}A + cos^{2}(30° - A) - sin A cos(30° - A) ). The student found that all these expressions equal the same constant. So if I can compute one of them, I can find the constant.Let me choose expression 2 because it's symmetric with A = 15°, so calculations might be simpler.Expression 2: ( sin^{2}15° + cos^{2}15° - sin15°cos15° )I know that ( sin^{2}A + cos^{2}A = 1 ), so that simplifies the first two terms:1 - ( sin15°cos15° )Now, ( sin15°cos15° ) can be rewritten using the double-angle identity:( sin2A = 2sin A cos A ), so ( sin A cos A = frac{1}{2}sin2A )Therefore, ( sin15°cos15° = frac{1}{2}sin30° )We know that ( sin30° = frac{1}{2} ), so:( sin15°cos15° = frac{1}{2} times frac{1}{2} = frac{1}{4} )Thus, expression 2 becomes:1 - ( frac{1}{4} ) = ( frac{3}{4} )So the constant is ( frac{3}{4} ).Now, to generalize this into a trigonometric identity. From the pattern observed, the identity would be:( sin^{2}A + cos^{2}(30° - A) - sin A cos(30° - A) = frac{3}{4} )I need to prove this identity. Let's start by expanding the terms.First, ( sin^{2}A + cos^{2}(30° - A) ). I know that ( cos(30° - A) ) can be expanded using the cosine of difference identity:( cos(30° - A) = cos30°cos A + sin30°sin A )We know that ( cos30° = frac{sqrt{3}}{2} ) and ( sin30° = frac{1}{2} ), so:( cos(30° - A) = frac{sqrt{3}}{2}cos A + frac{1}{2}sin A )Now, squaring this:( cos^{2}(30° - A) = left( frac{sqrt{3}}{2}cos A + frac{1}{2}sin A right)^2 )Expanding this:( left( frac{sqrt{3}}{2}cos A right)^2 + 2 times frac{sqrt{3}}{2}cos A times frac{1}{2}sin A + left( frac{1}{2}sin A right)^2 )Calculating each term:1. ( left( frac{sqrt{3}}{2}cos A right)^2 = frac{3}{4}cos^{2}A )2. ( 2 times frac{sqrt{3}}{2}cos A times frac{1}{2}sin A = frac{sqrt{3}}{2}sin A cos A )3. ( left( frac{1}{2}sin A right)^2 = frac{1}{4}sin^{2}A )So, adding them up:( frac{3}{4}cos^{2}A + frac{sqrt{3}}{2}sin A cos A + frac{1}{4}sin^{2}A )Now, let's look at the entire expression:( sin^{2}A + cos^{2}(30° - A) - sin A cos(30° - A) )We already have ( cos^{2}(30° - A) ) expanded. Let's substitute that in:( sin^{2}A + left( frac{3}{4}cos^{2}A + frac{sqrt{3}}{2}sin A cos A + frac{1}{4}sin^{2}A right) - sin A cos(30° - A) )Simplify the terms:Combine ( sin^{2}A ) and ( frac{1}{4}sin^{2}A ):( sin^{2}A + frac{1}{4}sin^{2}A = frac{5}{4}sin^{2}A )So now we have:( frac{5}{4}sin^{2}A + frac{3}{4}cos^{2}A + frac{sqrt{3}}{2}sin A cos A - sin A cos(30° - A) )Now, let's handle the last term: ( - sin A cos(30° - A) )Again, using the cosine of difference identity:( cos(30° - A) = frac{sqrt{3}}{2}cos A + frac{1}{2}sin A )So:( - sin A cos(30° - A) = - sin A left( frac{sqrt{3}}{2}cos A + frac{1}{2}sin A right) )Expanding this:( - frac{sqrt{3}}{2}sin A cos A - frac{1}{2}sin^{2}A )Now, substitute this back into the expression:( frac{5}{4}sin^{2}A + frac{3}{4}cos^{2}A + frac{sqrt{3}}{2}sin A cos A - frac{sqrt{3}}{2}sin A cos A - frac{1}{2}sin^{2}A )Simplify term by term:1. ( frac{5}{4}sin^{2}A - frac{1}{2}sin^{2}A = frac{5}{4}sin^{2}A - frac{2}{4}sin^{2}A = frac{3}{4}sin^{2}A )2. ( frac{sqrt{3}}{2}sin A cos A - frac{sqrt{3}}{2}sin A cos A = 0 )3. ( frac{3}{4}cos^{2}A ) remains as is.So, combining the remaining terms:( frac{3}{4}sin^{2}A + frac{3}{4}cos^{2}A )Factor out ( frac{3}{4} ):( frac{3}{4}(sin^{2}A + cos^{2}A) )We know that ( sin^{2}A + cos^{2}A = 1 ), so:( frac{3}{4} times 1 = frac{3}{4} )Thus, the identity holds true, and the constant is indeed ( frac{3}{4} ).**Final Answer**The constant is boxed{dfrac{3}{4}}.