(2ab/4a^2-9b^2-b/2a-3b):(1-2a-3b/2a+3b)

Решите задачу:

$( \frac{2ab}{4a^2-9b^2}- \frac{b}{2a-3b}):(1- \frac{2a-3b}{2a+3b})= \\ \\ =( \frac{2ab}{(2a-3b)(2a+3b)}- \frac{b(2a+3b)}{(2a-3b)(2a+3b)}):( \frac{2a+3b}{2a+3b} - \frac{2a-3b}{2a+3b})= \\ \\ =\frac{2ab-b(2a+3b)}{(2a-3b)(2a+3b)}:( \frac{2a+3b-2a+3b}{2a+3b})= \\ \\ =\frac{2ab-2ab-3b^2)}{(2a-3b)(2a+3b)}:( \frac{6b}{2a+3b})=\frac{(2ab-2ab-3b^2)}{(2a-3b)(2a+3b)}\cdot \frac{2a+3b}{6b}= \frac{-b}{2\cdot(2a-3b)}$