Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
P = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{3 + \sqrt x }}\\
= \dfrac{{15\sqrt x - 11}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} - \dfrac{{3\sqrt x - 2}}{{\sqrt x - 1}} - \dfrac{{2\sqrt x + 3}}{{3 + \sqrt x }}\\
= \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3x + 7\sqrt x - 6} \right) - \left( {2x + \sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {5\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow P = \dfrac{{2 - 5.3}}{{3 + 3}} = \dfrac{{2 - 15}}{6} = \dfrac{{ - 13}}{6}\\
c)P = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\
\Leftrightarrow \sqrt x + 3 = 4 - 10\sqrt x \\
\Leftrightarrow 11\sqrt x = 1\\
\Leftrightarrow \sqrt x = \dfrac{1}{{11}}\\
\Leftrightarrow x = \dfrac{1}{{121}}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{{121}}
\end{array}$
