Question

А,Б,В решить надо. Заранее спасибо за ответ

Answer 1

$displaystyle 4)(frac{2ab}{(a-b)(a+b)} +frac{(a-b)}{2(a+b)} )*frac{2a}{a+b} +frac{b}{a-b} =\\=(frac{4ab}{2(a-b)(a+b)} +frac{a^2-2ab+b^2}{2(a+b)(a-b)} )*frac{2a}{a+b} +frac{b}{a-b} =\\=frac{(a+b)^2}{2(a-b)(a+b)} *frac{2a}{a+b} +frac{b}{a-b} =\\=frac{a}{a-b} +frac{b}{a-b} =\\=frac{a+b}{a-b}$

$displaystyle 5)frac{y}{x-y} -frac{x(x-y)(x+y)}{x^2+y^2} *(frac{x}{(x-y)(x-y)}-frac{y}{(x-y)(x+y)}) =\\=frac{y}{x-y} -frac{x(x-y)(x+y)}{x^2+y^2} *(frac{x(x+y)}{(x-y)(x-y)(x+y)}-frac{y(x-y)}{(x-y)(x+y)(x-y)}) =\\=frac{y}{x-y} -frac{x(x-y)(x+y)}{x^2+y^2} *frac{x^2+y^2}{(x-y)(x-y)(x+y)} =\\=frac{y}{x-y} -frac{x}{x-y} =\\=-1$

$displaystyle 6)(frac{x}{(x-6)(x+6)} -frac{x-6}{x(x+6)} )*frac{x(x+6)}{2(x-3)} +frac{x}{6-x} =\\=(frac{x^2}{x(x-6)(x+6)} -frac{(x-6)^2}{x(x+6)(x-6)} )*frac{x(x+6)}{2(x-3)} +frac{x}{6-x} =\\=frac{12(x-3)}{x(x-6)(x+6)} *frac{x(x+6)}{2(x-3)} +frac{x}{6-x} =\\=frac{6}{x-6} +frac{x}{6-x} =\\=frac{6-x}{x-6} =-1$