Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4;x \ne 9
\end{array} \right.\\
b)A = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
+ \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - \left( {x - 9} \right) + 2x - 4\sqrt x + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
c)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}}\\
A \in Z\\
\Rightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in \left\{ { - 4; - 2; - 1;1;2;4} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 1;1;2;4;5;7} \right\}\\
Do:\sqrt x \ge 0\\
\Rightarrow \sqrt x \in \left\{ {1;2;4;5;7} \right\}\\
\Rightarrow x \in \left\{ {1;4;16;25;49} \right\}\\
Do:x \ne 4;x \ne 9\\
\Rightarrow x \in \left\{ {1;16;25;49} \right\}\\
d)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} - 1 < 0\\
\Rightarrow \dfrac{{\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\Rightarrow \dfrac{4}{{\sqrt x - 3}} < 0\\
\Rightarrow \sqrt x - 3 < 0\\
\Rightarrow \sqrt x < 3\\
\Rightarrow x < 9\\
\text{Vậy}\,0 \le x < 9;x \ne 4
\end{array}$
