Упростите выражение 2y^2 - 5xy/ x^2-4y^2 - x/2y - x - y/x+2y y+12/8y+32 - y+4/8y-32 +...

Решите задачу:

$\frac{2y^2-5xy}{x^2-4y^2}-\frac{x}{2y-x}-\frac{y}{x+2y}=\frac{2y^2-5xy}{(x-2y)(x+2y)}+\frac{x}{x-2y}-\frac{y}{x+2y}=\frac{2y^2-5xy}{(x-2y)(x+2y)}+\frac{x(x+2y)}{(x-2y)(x+2y)}-\frac{y(x-2y)}{(x-2y)(x+2y)}=\frac{2y^2-5xy+x^2+2xy+2y^2-xy}{(x-2y)(x+2y)}=\frac{4y^2-4xy+x^2}{(x-2y)(x+2y)}=\frac{(2y-x)^2}{-(2y-x)(x+2y)}=\frac{2y-x}{-(x+2y)}=\frac{x-2y}{x+2y}$

$\frac{y+12}{8y+32}-\frac{y+4}{8y-32}+\frac{9}{y^2-16}=\frac{y+12}{8(y+4)}-\frac{y+4}{8(y-4)}+\frac{9}{(y-4)(y+4)}=\frac{(y+12)(y-4)}{8(y+4)(y-4)}-\frac{(y+4)^2}{8(y-4)(y+4)}+\frac{9*8}{8(y-4)(y+4)}=\frac{y^2+8y-36-y^2-8y-16+72}{8(y+4)(y-4)}=\frac{20}{8(y^2-16)}=\frac{5}{2y^2-32}$