Упростить выражение x^2+3x/(x-3)^2 : ( 3/x+3 + x^2+9/x^2-9 - 3/3-x )

Решите задачу:

$\frac{x^2 + 3x}{(x-3)^2} : ( \frac{3}{x+3} + \frac{x^2+9}{x^2 - 9} - \frac{3}{3-x})= \\ \\ = \frac{x(x + 3)}{(x-3)^2} : ( \frac{3}{x+3} + \frac{x^2+9}{(x-3)(x+3)} - \frac{3}{-(x-3)})= \\ \\ =\frac{x(x + 3)}{(x-3)^2} : ( \frac{3(x-3)}{(x+3)(x-3)} + \frac{x^2+9}{(x-3)(x+3)} + \frac{3(x+3)}{(x-3)(x+3)} ) = \\ \\ = \frac{x(x + 3)}{(x-3)^2} : \frac{3x - 9 +x^2+9 +3x+9}{(x-3)(x+3)} = \\ \\$
$=\frac{x(x + 3)}{(x-3)^2} : \frac{x^2 + 6x+9}{(x-3)(x+3)} = \\ \\ = \frac{x(x + 3)}{(x-3)^2} : \frac{x^2 + 2*x*3 + 3^2}{(x-3)(x+3)} = \\ \\ = \frac{x(x+3)}{(x-3)^2} : \frac{(x+3)^2}{(x-3)(x+3)} = \\ \\ = \frac{x(x+3)}{(x-3)^2} : \frac{x+3}{x-3} = \frac{x(x+3)}{(x-3)^2}* \frac{x-3}{x+3}= \\ \\ =\frac{x * 1 * 1 }{(x-3) * 1 } = \frac{x}{x-3}$