Question

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

Answer 1

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

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