Rút gọn $P.$

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x  - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)

$\begin{array}{l}P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\\,\,\,\,\, = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \sqrt x \left( {\sqrt x  + 3} \right) - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{ - 3\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\\,\,\,\,\, = \dfrac{{ - 3\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{{ - 3}}{{\sqrt x  + 3}}.\end{array}$

Vậy \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)