answer:For the first problem, we have: begin{align*} left(1right)left(x-2yright)^{2}-2yleft(y-2xright) &= x^{2}-4xy+4y^{2}-2y^{2}+4xy &= x^{2}+2y^{2}-2y^{2}+4xy-4xy &= x^{2}+2y^{2} &= boxed{x^{2}+2y^{2}}. end{align*} For the second problem, we proceed as follows: begin{align*} (2)(m+1+frac{2m+3}{m+1})÷frac{m^{2}-4}{m+1} &= frac{(m+1)(m+1)+2m+3}{m+1}cdot frac{m+1}{(m+2)(m-2)} &= frac{m^{2}+2m+1+2m+3}{(m+2)(m-2)} &= frac{m^{2}+4m+4}{(m+2)(m-2)} &= frac{(m+2)^{2}}{(m+2)(m-2)} &= frac{m+2}{m-2} &= boxed{frac{m+2}{m-2}}. end{align*}

answer:Since the graph of y = f^{-1}(x) passes through the point (-2, 0), it means that f^{-1}(-2) = 0. Therefore, f(0) = -2. For the function y = f(x+5), when x = -5, y = f(-5+5) = f(0) = -2. Thus, the graph of y = f(x+5) passes through the point (-5, -2). Therefore, the correct answer is boxed{text{C: } (-5,-2)}.

answer:1. **Understanding the problem**: We need to find integers (n) such that: [ 5 text{ divides } (3n - 2) ] and, [ 7 text{ divides } (2n + 1) ] 2. **Simplifying the conditions**: - For the first condition (5 text{ divides } (3n - 2)), this can be rewritten as: [ 3n - 2 equiv 0 pmod{5} implies 3n equiv 2 pmod{5} ] - Finding the modular inverse of 3 modulo 5, we find that (3 times 2 equiv 1 pmod{5}), thus the modular multiplicative inverse of 3 is 2. Therefore: [ n equiv 2 times 2 pmod{5} implies n equiv 4 pmod{5} ] - Thus, we have: [ n equiv 4 pmod{5} ] 3. **Simplifying the second condition**: - For the second condition (7 text{ divides } (2n + 1)), this can be rewritten as: [ 2n + 1 equiv 0 pmod{7} implies 2n equiv -1 pmod{7} ] - Simplifying (-1 pmod{7}), we get (6): [ 2n equiv 6 pmod{7} ] - Finding the modular inverse of 2 modulo 7, we find that (2 times 4 equiv 1 pmod{7}), thus the modular multiplicative inverse of 2 is 4. Therefore: [ n equiv 6 times 4 pmod{7} implies n equiv 24 pmod{7} implies n equiv 3 pmod{7} ] - Thus, we have: [ n equiv 3 pmod{7} ] 4. **Using the Chinese Remainder Theorem**: Since 5 and 7 are coprime, we can apply the Chinese Remainder Theorem (CRT). According to CRT, there exists a unique solution modulo the product of the moduli (i.e., 35): [ begin{cases} n equiv 4 pmod{5} n equiv 3 pmod{7} end{cases} ] Let's denote (n) as (n = 5k + 4) for some integer (k). We substitute this into the second congruence: [ 5k + 4 equiv 3 pmod{7} ] Simplifying, we get: [ 5k + 4 - 3 equiv 0 pmod{7} implies 5k + 1 equiv 0 pmod{7} implies 5k equiv -1 pmod{7} implies 5k equiv 6 pmod{7} ] - Finding the modular inverse of 5 modulo 7, we find that (5 times 3 equiv 1 pmod{7}), thus the modular multiplicative inverse of 5 is 3. Therefore: [ k equiv 6 times 3 pmod{7} implies k equiv 18 pmod{7} implies k equiv 4 pmod{7} ] - Thus, we have: [ k = 7m + 4 quad text{for integer } m ] - Therefore: [ n = 5k + 4 = 5(7m + 4) + 4 = 35m + 20 + 4 = 35m + 24 ] Thus, all integers (n) that satisfy both conditions are given by: [ n = 35m + 24 quad text{for integer } m ] Conclusion. [boxed{n = 35m + 24, text{ where } m text{ is an integer}}]

answer:We begin by recognizing that cos^2x = 1 - sin^2x. Then the given function can be rewritten using this trigonometric identity: begin{align*} y &= -cos^2x + 2sin x + 2 &= -(1 - sin^2x) + 2sin x + 2 &= -1 + sin^2x + 2sin x + 2 &= sin^2x + 2sin x + 1 &= (sin x + 1)^2. end{align*} The expression (sin x + 1)^2 is a square of a real number, so it is non-negative for all x. The smallest value it can take is 0, which occurs when sin x + 1 = 0, that is, sin x = -1. Therefore, the minimum value of y is boxed{0}.