Задание А10 ,помогите пожалуйста

Решите задачу:

$(\frac{x+5}{(x-9)(x+9)}+ \frac{x+7}{(x-9)^2} )( \frac{x-9}{x+3} )^2+ \frac{7+x}{9+x}= \\=(\frac{(x+5)(x-9)}{(x-9)^2(x+9)}+ \frac{(x+7)(x+9)}{(x-9)^2(x+9)} ) \frac{(x-9)^2}{(x+3)^2} + \frac{7+x}{9+x}= \\ = \frac{(x+5)(x-9)+ (x+7)(x+9)}{(x-9)^2(x+9)} \frac{(x-9)^2}{(x+3)^2}+ \frac{7+x}{9+x}= \\ = \frac{x^2 -9x+5x-45+x^2+9x+7x+63}{x+9} \frac{1}{(x+3)^2}+ \frac{7+x}{9+x}= \\ = \frac{2x^2 +12x+18}{(x+9)(x+3)^2} + \frac{7+x}{9+x}=\frac{2(x^2 +6x+9)}{(x+9)(x+3)^2} + \frac{7+x}{9+x}=$
$=\frac{2(x +3)^2}{(x+9)(x+3)^2} + \frac{7+x}{9+x}=\frac{2}{(x+9)} + \frac{7+x}{9+x}=\frac{2 +7+x}{9+x}=\frac{9+x}{9+x}=1$