3) \(\dfrac{4}{x+3}-\dfrac{5}{x-3}+\dfrac{4x+36}{9-x^2}\)

Область допустимых значений: \(x\ne -3\) и \(x\ne 3\). Используем равенство \(9-x^2=-(x-3)(x+3)\).

\(\dfrac{4}{x+3}-\dfrac{5}{x-3}+\dfrac{4x+36}{9-x^2}=\dfrac{4(x-3)-5(x+3)-(4x+36)}{(x+3)(x-3)}=\dfrac{-5x-63}{x^2-9}.\)

Ответ: \(\dfrac{-5x-63}{x^2-9}\), \(x\ne\pm3\).