60 баллов!!!!!!!!!!!!!!!!! 2 примера... решите плиз..))

Решите задачу:

$(2a-\frac{a^2-1}{1+a}):\frac{a^2+2a+1}{a-2}+\frac{2}{a+1}=(2a-\frac{(a-1)(a+1)}{1+a}):\frac{(a+1)^2}{a-2}+\frac{2}{a+1}=\\\ =(2a-a+1):\frac{(a+1)^2}{a-2}+\frac{2}{a+1}=(a+1)\cdot\frac{a-2}{(a+1)^2}+\frac{2}{a+1}=\\\ =\frac{a-2}{a+1}+\frac{2}{a+1}=\frac{a-2+2}{a+1}=\frac{a}{a+1}$

$(\frac{1}{y^2+3y}-\frac{1}{y+3}+1)\cdot(\frac{1}{y^2-1}-\frac{1}{(y+1)^3})-\frac{y^2}{y^2-1}=\\\ =(\frac{1}{y(y+3)}-\frac{1}{y+3}+1)\cdot(\frac{1}{(y-1)(y+1)}-\frac{1}{(y+1)^3})-\frac{y^2}{y^2-1}=\\\ =\frac{1-y+y^2+3y}{y(y+3)}\cdot\frac{(y+1)^2+(y-1)}{(y-1)(y+1)^3}-\frac{y^2}{y^2-1}=\\\ =\frac{y^2+2y+1}{y^2+3y}\cdot\frac{y^2+2y+1+y-1}{(y-1)(y+1)^3}-\frac{y^2}{y^2-1}=\\\ =\frac{(y+1)^2}{y^2+3y}\cdot\frac{y^2+3y}{(y-1)(y+1)^3}-\frac{y^2}{y^2-1}=\\\$
$=\frac{1}{(y-1)(y+1)}-\frac{y^2}{y^2-1}=\frac{1}{y^2-1}-\frac{y^2}{y^2-1}=\frac{1-y^2}{y^2-1}=\\\ =-\frac{y^2-1}{y^2-1}=-1$