Question

решите задание 1 варианта пожалуйста!!! Дам 50 баллов

Answer 1

ВI.
1.
$1) \frac{x^2}{x^2-y^2} * \frac{x-y}{x} = \frac{x^2(x-y)}{(x-y)(x+y)*x} = \frac{x*x(x-y)}{(x+y) * x(x-y)} = \frac{x}{x+y} \\ \\ 2) \frac{2x}{x-a} - \frac{2a}{x+a} = \frac{2x(x+a) -2a(x-a)}{(x-a)(x+a)} = \frac{2x^2 +2ax -2ax +2a^2}{(x-a)(x+a)} = \frac{2x^2+2a^2}{x^2-a^2}$

$3) c+\frac{c^2}{c+1} = \frac{c(c+1)+c^2}{c+1} = \frac{c^2+c+c^2}{c+1} = \frac{2c^2+c}{c+1} \\ \\ 4) \frac{a}{3a+3b} : \frac{a^2}{a^2-b^2} = \frac{a}{3(a+b)} * \frac{(a-b)(a+b)}{a^2} = \frac{(a-b) * a(a+b)}{3a *a(a+b)} = \frac{a-b}{3a} = \\ \\ = \frac{a}{3a} - \frac{b}{3a} = \frac{1}{3} - \frac{b}{3a} = \frac{1}{3} (1- \frac{b}{a} )$

2.
$1) (\frac{y}{y-x} - \frac{y-x}{y} ) * \frac{y-x}{x} = \frac{y}{y-x} * \frac{y-x}{x} - \frac{y-x}{y} * \frac{y-x}{x} = \frac{y}{x} - \frac{(y-x)^2}{xy} = \\ \\ = \frac{y*y - (y-x)^2}{xy} = \frac{y^2-y^2+2xy-x^2}{xy} = \frac{2xy-x^2}{xy} = \frac{x(2y-x)}{xy} = \frac{2y-x}{x} = \\ \\ = \frac{2y}{x} - \frac{x}{x} = \frac{2y}{x} - 1$
$2) \frac{2x-4}{x^2+12x+36} : \frac{8x-16}{x^2-36} = \frac{2(x-2)}{x^2+2*x*6+6^2} : \frac{8(x-2)}{x^2-6^2} = \\ \\ = \frac{2(x-2)}{(x+6)^2} * \frac{(x-6)(x+6)}{8(x-2)} = \frac{(x-6) * 2(x-2)(x+6)}{4(x+6)*2(x-2)(x+6)} = \\ \\ = \frac{x-6}{4(x+6)} = \frac{x-6}{4x+24}$