Đáp án: $P = \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
P = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3x + 7\sqrt x - 6} \right) - \left( {2x + \sqrt x - 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - \left( {5\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}
\end{array}$
