Решите плиз   (6/(y^2-9)+1/(3-y))*((y^2+6*y+9)/5)  

[latex](\frac {6}{y^2-9}+\frac {1}{3-y}) \cdot (\frac {y^2+6y+9}{5}) \\ \\ \\ \\ 1) \ \frac {6}{y^2-9}+\frac {1}{3-y}=\frac {6}{(y-3)(y+3)}+\frac {1}{3-y}=\frac {6}{(y-3)(y+3)}-\frac {1}{y-3}= \\ \\ =\frac {6}{(y-3)(y+3)}-\frac {3+y}{(y-3)(y+3)}=\frac {6-(3+y)}{(y-3)(y+3)}=\frac {6-3-y}{(y-3)(y+3)}=\\ \\ =\frac {3-y}{(y-3)(y+3)}= -\frac {1}{3+y} \\ \\ \\ 2) -\frac {1}{3+y} \cdot \frac {y^2+6y+9}{5}=- \frac {1}{3+y} \cdot \frac {(y+3)^2}{5}=-\frac {1 \cdot (y+3)(y+3)}{(3+y) \cdot 5}=-\frac {y+3}{5} [/latex]     если в решении что-то неясно, пишите в лс