Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 4;x \ne 9\\
A = \dfrac{{2\sqrt x - 13}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 13}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x - 13 - {{\left( {\sqrt x - 3} \right)}^2} + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 13 - \left( {x - 6\sqrt x + 9} \right) + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 5\sqrt x - 24}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 8} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 8}}{{\sqrt x - 2}}\\
b)A < 1\\
\Rightarrow \dfrac{{\sqrt x + 8}}{{\sqrt x - 2}} < 1\\
\Rightarrow \dfrac{{\sqrt x + 8 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\\
\Rightarrow \dfrac{{10}}{{\sqrt x - 2}} < 0\\
\Rightarrow \sqrt x - 2 < 0\\
\Rightarrow \sqrt x < 2\\
\Rightarrow x < 4\\
\text{Vậy}\,0 \le x < 4\\
c)A = 6\\
\Rightarrow \dfrac{{\sqrt x + 8}}{{\sqrt x - 2}} = 6\\
\Rightarrow \sqrt x + 8 = 6\sqrt x - 12\\
\Rightarrow 5\sqrt x = 20\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\left( {tmdk} \right)\\
\text{Vậy}\,x = 16
\end{array}$
