Помогите упростить .

Решите задачу:

$\frac{4}{a-b} + \frac{9}{a+b} - \frac{8a}{a^2-b^2} = \frac{4(a+b)+9(a-b)-8a}{a^2-b^2}=\frac{5a-5b}{a^2-b^2}=\frac{5(a-b)}{(a-b)(a+b)}=\frac{5}{a+b}$

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

$\frac{b^2}{a^2-2ab} : (\frac{2ab}{a^2-4b^2}- \frac{b}{a+2b} )= \frac{b^2}{a^2-2ab} : \frac{2ab-b(a-2b)}{a^2-4b^2}= \frac{b^2}{a^2-2ab} : \frac{2ab-ab+2b^2}{a^2-4b^2}= \\\ \frac{b^2(a-2b)(a+2b)}{a(a-2b)b(a+2b)}= \frac{b}{a}$