Question

помогите, срочно, пожалуйста

Answer 1

($frac{6}{y^{2}-9 }$+$frac{1}{3-y}$)*$frac{ y^{2}+6y+9}{5}$
$frac{6}{(y-3)(y+3)}$+$frac{1}{3-y}$
$frac{6}{(y-3)(y+3)}$+$frac{-y-3}{(y-3)(y+3)}$
$frac{6-y-3}{(y-3)(y+3)}$
$frac{3-y}{(y-3)(y+3)}$
$frac{-1}{y+3}$*$frac{ (y+3)^{2} }{5}$
-1*$frac{y+3}{5}$
-$frac{y+3}{5}$