Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 9;x \ne 4\\
x = \dfrac{1}{9}\left( {tmdk} \right)\\
\Rightarrow \sqrt x = \dfrac{1}{3}\\
\Rightarrow A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\dfrac{1}{3} + 1}}{{\dfrac{1}{3} - 3}} = - \dfrac{1}{2}\\
2)B = \left( {\dfrac{2}{{\sqrt x - 3}} - \dfrac{{2\sqrt x - 1}}{{x - \sqrt x - 6}}} \right):\dfrac{5}{{x - 4}}\\
= \dfrac{{2\left( {\sqrt x + 2} \right) - 2\sqrt x + 1}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{x - 4}}{5}\\
= \dfrac{{2\sqrt x + 4 - 2\sqrt x + 1}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{5}\\
= \dfrac{5}{{\sqrt x - 3}}.\dfrac{{\sqrt x - 2}}{5}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}\\
3)\\
Dkxd:x \ge 0;x \ne 9;x \ne 4\\
\dfrac{B}{A} = \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
\dfrac{B}{A} \le \dfrac{2}{3}\\
\Rightarrow \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} - \dfrac{2}{3} \le 0\\
\Rightarrow \dfrac{{3\sqrt x - 6 - 2\sqrt x - 2}}{{3\left( {\sqrt x + 1} \right)}} \le 0\\
\Rightarrow \sqrt x - 8 \le 0\left( {do:\sqrt x + 1 > 0} \right)\\
\Rightarrow \sqrt x \le 8\\
\Rightarrow x \le 64\\
Vay\,0 \le x \le 64;x \ne 9;x \ne 4
\end{array}$
