Второе задание. Алгебра.

Решите задачу:

$\frac{x^2+3x}{(x-3)^2}:\left(\frac3{x+3}+\frac{x^2+9}{x^2-9}-\frac3{3-x}\right)=\\=\frac{x^2+3x}{(x-3)^2}:\left(\frac3{x+3}+\frac{x^2+9}{(x-3)(x+3)}+\frac3{x-3}\right)=\\=\frac{x^2+3x}{(x-3)^2}:\left(\frac{3(x-3)+x^2+9+3(x+3)}{(x-3)(x+3)}\right)=\frac{x^2+3x}{(x-3)^2}:\left(\frac{3x-9+x^2+9+3x+9}{(x-3)(x+3)}\right)=\\=\frac{x(x+3)}{(x-3)^2}\cdot\frac{(x-3)(x+3)}{x^2+6x+9}=\frac{x(x+3)}{(x-3)^2}\cdot\frac{(x-3)(x+3)}{(x+3)^2}=\frac x{x-3}\\x=-299\\\frac{-299}{-299-3}=\frac{299}{302}$