Question

Помогите сделать номер 560

Answer 1

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