Вопрос в картинках...

Решите задачу:

$1)\; \; \frac{2x}{x^2-y^2} :( \frac{1}{x^2+2xy+y^2} - \frac{1}{y^2}-x^2)=\\\\=\frac{2x}{(x-y)(x+y)} : ( \frac{1}{(x+y)^2}-\frac{1}{y^2}-x^2 )= \\\\=\frac{2x}{(x-y)(x+y)}:\frac{y^2-(x+y)^2-x^2y^2(x+y)^2}{y^2(x+y)^2}=\\\\= \frac{2x}{(x-y)(x+y)}\cdot \frac{y^2(x+y)^2}{y^2-(x+y)^2-x^2y^2(x+y)^2}=\frac{2xy^2(x+y)}{(x-y)\cdot (y^2-(x+y)^2-x^2y^2(x+y)^2)}$

$2)\; \; ...= \frac{2x}{(x-y)(x+y)}:( \frac{1}{(x+y)^2}- \frac{1}{y^2-x^2} )=\\\\= \frac{2x}{(x-y)(x+y)}:( \frac{1}{(x+y)^2}-\frac{1}{(y-x)(y+x)})=\\\\=\frac{2x}{(x-y)(x+y)}:(\frac{1}{(x+y)^2}+\frac{1}{(x-y)(x+y)})= \frac{2x}{(x-y)(x+y)} : \frac{x-y+x+y}{(x+y)^2(x-y)} =\\\\= \frac{2x}{(x-y)(x+y)} \cdot \frac{(x+y)^2(x-y)}{2x}=x+y$