Đáp án:
$\begin{array}{l}
P = \frac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \frac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
Đkxđ:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
P = \frac{{15\sqrt x - 11}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \frac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \frac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{15\sqrt x - 11 - \left( {3x + 9\sqrt x - 2\sqrt x - 6} \right) - \left( {2x - 2\sqrt x + 3\sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( { - 5\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{ - 5\sqrt x + 2}}{{\sqrt x + 3}}\\
b)x \ge 0;x \ne 1\\
P = \frac{1}{2}\\
\Rightarrow \frac{{ - 5\sqrt x + 2}}{{\sqrt x + 3}} = \frac{1}{2}\\
\Rightarrow - 10\sqrt x + 4 = \sqrt x + 3\\
\Rightarrow 11\sqrt x = 1\\
\Rightarrow \sqrt x = \frac{1}{{11}}\\
\Rightarrow x = \frac{1}{{121}}\left( {tmdk} \right)\\
Vậy\,x = \frac{1}{{121}}\,thì\,P = \frac{1}{2}
\end{array}$
