Докажите тождество: (y^2+9/27-3y^2 + y/3y+9 - 3/y^2-3y): (3y+9)^2/3y^2-y^3= y/9y+27

[latex](frac{y^2+9}{27-3y^2}+frac{y}{3y+9}-frac{3}{y^2-3y}):frac{(3y+9)^2}{3y^2-y^3}[/latex][latex]1) frac{y^2+9}{27-3y^2}+frac{y}{3y+9}-frac{3}{y^2-3y}=frac{y^2+9}{-3(y-3)(y+3)}+frac{y}{3(y+3)}-frac{3}{y(y-3)}= \ \ =frac{y}{3(y+3)}-frac{y^2+9}{3(y-3)(y+3)}-frac{3}{y(y-3)}= frac{y^2(y-3)-y(y^2+9)-9(y+3)}{3y(y-3)(y+3)} = \ \ =frac{(y^3-3y^2)-(y^3+9y)-(9y+27)}{3y(y-3)(y+3)} =frac{y^3-3y^2-y^3-9y-9y-27}{3y(y-3)(y+3)} = \ \ =frac{-3y^2-18y-27}{3y(y-3)(y+3)} =frac{-3(y+3)^2}{3y(y-3)(y+3)} =-frac{y+3}{y(y-3)}[/latex][latex]2) -frac{y+3}{y(y-3)}:frac{(3y+9)^2}{3y^2-y^3}=-frac{y+3}{y(y-3)}:(-frac{9(y+3)^2}{y^2(y-3)})= \ \ =frac{y+3}{y(y-3)}*frac{y^2(y-3)}{9(y+3)^2}=frac{y}{9(y+3)}=frac{y}{9y+27}[/latex]ответ: тождество верно