Question

Срочно, помогите прошу

Answer 1

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