Решите, пожалуйста срочно

Решите задачу:

$\left (\frac{2x}{2x+y} - \frac{4x^2}{4x^2+4xy+y^2} \right ):\left ( \frac{2x}{4x^2-y^2} + \frac{1}{y-2x} \right )=\\\\=\left ( \frac{2x}{2x+y} - \frac{4x^2}{(2x+y)^2} \right ):\left ( \frac{2x}{(2x-y)(2x+y)} - \frac{1}{2x-y} \right )=\\\\= \frac{2x(2x+y)-4x^2}{(2x+y)^2} : \frac{2x-(2x+y)}{(2x-y)(2x+y)} = \frac{2xy}{(2x+y)^2} \cdot \frac{(2x-y)(2x+y)}{-y} =\\\\= -\frac{2x(2x-y)}{2x+y} = \frac{2x(y-2x)}{2x+y}$

$2)\quad \sqrt{(2-\sqrt5)^2} + \sqrt{(3-\sqrt5)^2} =|2-\sqrt5|+|3-\sqrt5|=\\\\=-(2-\sqrt5)+(3-\sqrt5)=-2+\sqrt5+3-\sqrt5=-2+3=1\\\\P.S.\; \; \; \sqrt5\approx 2,24\ \textgreater \ 2\; \; \Rightarrow \; \; 2-\sqrt5\ \textless \ 0\; \; \Rightarrow \; \; |2-\sqrt5|=-(2-\sqrt5)$