\[\begin{array}{l}
A = \frac{{2\sqrt x + 1}}{{\sqrt x - 3}} + \frac{{\sqrt x + 3}}{{2 - \sqrt x }} - \frac{{10 - \sqrt x }}{{x - 5\sqrt x + 6}}\\
dk:x \ge 0;x \ne 4;x \ne 9\\
= \frac{{\left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} - \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} - \frac{{10 - \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{2x - 4\sqrt x + \sqrt x - 2 - x + 9 - 10 + \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{x - 2\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x + 1}}{{\sqrt x - 2}}
\end{array}\]

Vậy  với \[x \ge 0;x \ne 4;x \ne 9\] thì \[A = \frac{{\sqrt x + 1}}{{\sqrt x - 2}}\]