Question

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

Answer 1

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

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

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