Упростите выражение

Решите задачу:

$\frac{1,5x}{x+y} + \frac{x-y}{x^2+2xy}:( \frac{y}{x^2-2xy}- \frac{2x+y}{x^2-4y^2} )= \\\ =\frac{1,5x}{x+y} + \frac{x-y}{x(x+2y)}:( \frac{y}{x(x-2y)}- \frac{2x+y}{(x-2y)(x+2y)} )= \\\ =\frac{1,5x}{x+y} + \frac{x-y}{x(x+2y)}:\frac{y(x+2y)-(2x+y)x}{x(x-2y)(x+2y)}= \\\ =\frac{1,5x}{x+y} + \frac{x-y}{x(x+2y)}:\frac{xy+2y^2-2x^2-xy}{x(x-2y)(x+2y)}= \\\ =\frac{1,5x}{x+y} + \frac{x-y}{x(x+2y)}:\frac{2(y-x)(y+x)}{x(x-2y)(x+2y)}=$
$=\frac{1,5x}{x+y} + \frac{x-y}{x(x+2y)}\cdot\frac{x(x-2y)(x+2y)}{2(y-x)(y+x)}= \\\ =\frac{1,5x}{x+y} -\frac{(x-y)x(x-2y)(x+2y)}{2(x-y)(y+x)x(x+2y)}= \\\ =\frac{3x}{2(x+y)} -\frac{x-2y}{2(y+x)}=\frac{3x-x+2y}{2(x+y)}=\frac{2x+2y}{2x+2y}=1$