Đáp án:
$\begin{array}{l}
Đkxđ:\left\{ \begin{array}{l}
x + 2\sqrt x - 3 \ne 0\\
1 - \sqrt x \ne 0\\
\sqrt x + 3 \ne 0\\
x \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right) \ne 0\\
x \ne 1\\
x \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
P = \left( {\frac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \frac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}} \right)\\
= \frac{{15\sqrt x - 11}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \frac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \frac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{15\sqrt x - 11 - 3x - 9\sqrt x + 2\sqrt x + 6 - 2x + 2\sqrt x - 3\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{\left( { - 5\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{ - 5\sqrt x + 2}}{{\sqrt x + 3}}
\end{array}$
