Ответ:

[tex]-\frac{1}{a}[/tex]

Объяснение:

[tex](\frac{2}{3a+b} - \frac{1}{3a-b} - \frac{4b}{b^{2}-9a^{2} }) * (\frac{b}{a} - 3) = (\frac{2}{3a+b} - \frac{1}{3a-b} - \frac{4b}{(b-3a)(b+3a)}) * \frac{b-3a}{a} = (\frac{2}{3a+b} - \frac{1}{3a-b} - \frac{4b}{-(3a-b)(b+3a)}) * \frac{-(3a-b)}{a} = (\frac{2}{3a+b} - \frac{1}{3a-b} + \frac{4b}{(3a-b)(b+3a)}) * \frac{-(3a-b)}{a} = \frac{2(3a-b)-(3a+b)+4b}{(3a-b)(3a+b)} * \frac{-(3a-b)}{a} = \frac{6a-2b-3a-b+4b}{(3a-b)(3a+b)} * \frac{-(3a-b)}{a} = \frac{3a-2b-b+4b}{(3a-b)(3a+b)} * \frac{-(3a-b)}{a} =[/tex]

[tex]\frac{3a+b}{(3a-b)(3a+b)} * \frac{-(3a-b)}{a} = \frac{1}{(3a-b)} * \frac{-(3a-b)}{a} = \frac{-1}{a} = -\frac{1}{a}[/tex]