Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 4;x\# 9\\
A = \left( {\dfrac{{x - 9\sqrt x - 7}}{{x - 9}} + \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} - \dfrac{1}{{\sqrt x + 3}}} \right)\\
:\left( {\dfrac{1}{{\sqrt x - 3}} + 1} \right)\\
= \dfrac{{x - 9\sqrt x - 7 + \left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right) - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
:\dfrac{{1 + \sqrt x - 3}}{{\sqrt x - 3}}\\
= \dfrac{{x - 9\sqrt x - 7 + x + 5\sqrt x + 6 - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
.\dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
= \dfrac{{2x - 5\sqrt x + 2}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}}\\
2)A = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}}\\
= \dfrac{{2\sqrt x + 6 - 7}}{{\sqrt x + 3}}\\
= 2 - \dfrac{7}{{\sqrt x + 3}}\\
A \in Z\\
\Leftrightarrow \dfrac{7}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 = 7\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16\\
3)A < \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}} < \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{6\sqrt x - 3 - 2\sqrt x - 6}}{{3\left( {\sqrt x + 3} \right)}} < 0\\
\Leftrightarrow 4\sqrt x - 9 < 0\\
\Leftrightarrow \sqrt x < \dfrac{9}{4}\\
\Leftrightarrow x < \dfrac{{81}}{{16}}\\
Vậy\,0 \le x \le \dfrac{{81}}{{16}};x\# 4\\
4)x = 17 - 12\sqrt 2 \\
= 9 - 2.3.2\sqrt 2 + 8\\
= {\left( {3 - 2\sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = 3 - 2\sqrt 2 \\
A = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}}\\
= \dfrac{{2.\left( {3 - 2\sqrt 2 } \right) - 1}}{{3 - 2\sqrt 2 + 3}}\\
= \dfrac{{5 - 4\sqrt 2 }}{{6 - 2\sqrt 2 }}\\
= \dfrac{{\left( {5 - 4\sqrt 2 } \right)\left( {3 + \sqrt 2 } \right)}}{{2.\left( {9 - 2} \right)}}\\
= \dfrac{{7 - 7\sqrt 2 }}{{2.7}}\\
= \dfrac{{1 - \sqrt 2 }}{2}
\end{array}$
