Rút gọn P ta được:

Ta có: \(P = \left( {\dfrac{x}{{x + 3}} - \dfrac{2}{{x - 3}} - \dfrac{{{x^2} - 1}}{{9 - x{}^2}}} \right):\left( {2 - \dfrac{{x + 5}}{{x + 3}}} \right)\).

ĐK: \(x \ne 3;x \ne  - 3\); \(x\ne -1\).

\(P = \left( {\dfrac{x}{{x + 3}} - \dfrac{2}{{x - 3}} - \dfrac{{{x^2} - 1}}{{9 - x{}^2}}} \right):\left( {2 - \dfrac{{x + 5}}{{x + 3}}} \right)\)\( = \left( {\dfrac{x}{{x + 3}} - \dfrac{2}{{x - 3}} + \dfrac{{{x^2} - 1}}{{(x - 3)(x + 3)}}} \right):\left( {\dfrac{{2x + 6 - x - 5}}{{x + 3}}} \right)\)\( = \left( {\dfrac{{x(x - 3) - 2(x + 3) + {x^2} - 1}}{{(x + 3)(x - 3)}}} \right):\dfrac{{x + 1}}{{x + 3}}\)\(= \dfrac{{2{x^2} - 5x - 7}}{{(x + 3)(x - 3)}}.\dfrac{{x + 3}}{{x + 1}}\)\( = \dfrac{{2{x^2} + 2x - 7x - 7}}{{(x + 3)(x - 3)}}.\dfrac{{x + 3}}{{x + 1}}\)\(= \dfrac{{(2x - 7)(x + 1)}}{{(x + 3)(x - 3)}}.\dfrac{{x + 3}}{{x + 1}}\)\(= \dfrac{{2x - 7}}{{x - 3}}\)

Vậy \(P = \dfrac{{2x - 7}}{{x - 3}}\) với \(x \ne 3;x \ne  - 3;x\ne -1.\)