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Решите задачу:

$( \frac{a-3b}{a^2+3ab} - \frac{a+3b}{a^2-3ab} ): \frac{4b^2}{9b^2-a^2}= ( \frac{a-3b}{a(a+3b)} - \frac{a+3b}{a(a-3b)} )* \frac{9b^2-a^2}{4b^2}= \\ \\ \frac{(a-3b)^2-(a+3b)^2}{a(a+3b)(a-3b)}* \frac{(3b-a)(3b+a)}{4b^2}= \frac{a^2-6ab+9b^2-(a^2+6ab+9b^2)}{a(a-3b)}* \frac{(3b-a)}{4b^2}= \\ \\ \frac{a^2-6ab+9b^2-a^2-6ab-9b^2}{a(a-3b)}* \frac{(3b-a)}{4b^2}= \frac{-12ab}{a(a-3b)}* \frac{-(a-3b)}{4b^2}= \frac{12b}{4b^2} = \frac{3}{b}$