Đáp án: $P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 4;x\# 9\\
P = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - \left( {x - 9} \right) + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}
\end{array}$
