Ответ:

4) \(\frac{b-2}{b^2+6b+9} - \frac{b}{b^2-9} = \frac{b-2}{(b+3)^2} - \frac{b}{(b-3)(b+3)} = \frac{(b-2)(b-3)}{(b+3)^2(b-3)} - \frac{b(b+3)}{(b-3)(b+3)^2} = \frac{b^2-5b+6-b^2-3b}{(b+3)^2(b-3)} = \frac{-8b+6}{(b+3)^2(b-3)}\). Итоговый ответ: \(\frac{-8b+6}{(b+3)^2(b-3)}\)