Aaaaa!!!Помогитееееее!!!

Решите задачу:

$\frac{x+2}{ x^{2} -2x+1}* \frac{3x-3}{ x^{2} -4} - \frac{3}{x-2}= \frac{x+2}{(x-1) ^{2} }* \frac{3(x-1)}{(x+2)(x-2)}- \frac{3}{x-2} = \frac{3}{(x-1)(x-2)}- \frac{3}{x-2}$ $= \frac{3-3(x-1)}{(x-1)(x-2)}= \frac{3-3x+3}{(x-1)(x-2)}= \frac{6-3x}{(x-1)(x-2)}= \frac{3(2-x)}{(x-1)(x-2)} =- \frac{3}{x-1}= \frac{3}{1-x}\\\\ \frac{3}{1-(-1,5)}= \frac{3}{1+1,5}= \frac{3}{2,5}=1,2\\\\\\( \frac{ y^{2}-3y }{ y^{2}-6y+9 }- \frac{3y+9}{ y^{2}-9 } )*(1- \frac{3}{y})=( \frac{y(y-3)}{(y-3) ^{2} }- \frac{3(y+3)}{(y-3)(y+3)})* \frac{y-3}{y} =( \frac{y}{y-3} -$ $- \frac{3}{y-3})* \frac{y-3}{y} = \frac{y-3}{y-3} * \frac{y-3}{y} =1* \frac{y-3}{y}= \frac{y-3}{y} \\\\ \frac{-3,6-3}{-3,6} = \frac{-6,6}{-3,6} =1 \frac{5}{6}$