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