Помогите решить пример

Question 1

Question 2

Решите задачу:

$( \frac{x-2y}{x^2 +2xy} - \frac{x+2y}{x^2 -2xy} ) : \frac{4y^2}{4y^2 -x^2} = \\ \\ =( \frac{x-2y}{x(x+2y) } - \frac{x+2y}{x(x-2y)} ) : \frac{4y^2}{4y^2 -x^2} = \\ \\ = ( \frac{(x-2y)(x-2y) - (x+2y)(x+2y)}{x(x+2y)(x-2y)} ) : \frac{4y^2}{4y^2 -x^2} = \\ \\ =( \frac{(x-2y)^2 - (x+2y)^2}{x(x+2y)(x-2y) } ) : \frac{4y^2}{- 1 (x^2 - 4y^2)} = \\ \\ = \frac{x^2 - 2*x*2y +(2y)^2 - (x^2 + 2*x*2y + (2y)^2 )}{x(x^2 - (2y)^2 )} * ( - \frac{x^2-4y^2}{4y^2} ) = \\ \\$
$= \frac{x^2 - 2*x*2y +(2y)^2 - (x^2 + 2*x*2y + (2y)^2 )}{x(x^2 - (2y)^2 )} * ( - \frac{x^2-4y^2}{4y^2} ) = \\ \\ = \frac{x^2 - 4xy +4y^2 - x^2 -4xy -4y^2}{x(x^2 -4y^2)} * (- \frac{x^2 -4y^2}{4y^2} )= \\ \\ = \frac{-8xy}{x(x^2 -4y^2) } * (- \frac{x^2 -4y^2}{4y^2} )= \\ \\ = - \frac{8y }{x^2 -4y^2} * ( - \frac{x^2 - 4y^2}{4y^2} ) = \\ \\ = \frac{2*4y *(x^2 -4y^2)}{y*4y*(x^2 -4y^2)} = \frac{2}{y}$