Срочняк надо :3срочняк надо :3срочняк надо :3

Решите задачу:

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