Помогите не сложно! !!!

Решите задачу:

$( \frac{8x}{x-2}+2x): \frac{3(x+2)}{6(x-2)}=( \frac{8x+2x^2-4x}{x-2} )* \frac{2(x-2)}{x+2}= \frac{2x(x+2)}{x-2}=2*2x=4x\\\\( \frac{d+9}{d}- \frac{c-9}{c})* \frac{cd}{d^2-c^2}=( \frac{cd+9c-cd+9d}{cd} )*\frac{cd}{d^2-c^2}= \frac{9(c+d)}{cd}*\frac{cd}{(d-c)(d+c)}= \frac{9}{d-c}\\\\ (\frac{3-x}{x^3}* \frac{xy-x}{3-x})+ \frac{3-y}{3x^2}=( \frac{3-x}{x^3}* \frac{x(y-1)}{3-x})+ \frac{3-y}{3x^2}= \frac{y-1}{x^2}+ \frac{3-y}{3x^2}= \frac{3y-3+3-y}{3x^2}=\\ \frac{2y}{3x^2}$