Question

помогите пожалуйста с 6 номером все на фото

Answer 1

$( \frac{9}{y^2-9} + \frac{3}{(3-y)^2}) : \frac{6}{(y-3)^2}+ \frac{1-2y}{3+y} =\\( \frac{9}{(y-3)(y+3)} + \frac{3}{(y-3)(y-3)} ) : \frac{6}{(y-3)(y-3)} + \frac{1-2y}{3+y}=\\( \frac{9(y-3)+3(y+3)}{(y-3)(y+3)(y-3)} ) : \frac{6}{(y-3)(y-3)} + \frac{1-2y}{3+y}=\\( \frac{9y-27+3y+9}{(y-3)^2(y+3)} ) : \frac{6}{(y-3)^2}+ \frac{1-2y}{3+y}=\\\frac{12y-18}{(y-3)^2(y+3)} * \frac{(y-3)^2}{6} + \frac{1-2y}{3+y}=\\ \frac{6(2y-3)(y-3)^2}{(y-3)^2(y+3)6} + \frac{1-2y}{3+y}=\\ \frac{2y-3}{y+3} + \frac{1-2y}{y+3}=$
$\frac{2y-3+1-2y}{y+3} = \frac{-2}{y+3}$
Если $y=-3.4$, то $\frac{-2}{-3.4+3} = \frac{-2}{-0.4} =5$