Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 4\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} - \dfrac{5}{{x + \sqrt x - 6}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - 5 - \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 - 5 - \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 12}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}\\
b)x = 6 + 4\sqrt 2 \left( {tmdk} \right)\\
x = 4 + 2.2.\sqrt 2 + 2\\
x = {\left( {2 + \sqrt 2 } \right)^2}\\
\Rightarrow \sqrt x = 2 + \sqrt 2 \\
\Rightarrow A = \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt 2 - 2}}{{\sqrt 2 }}\\
= 1 - \sqrt 2
\end{array}$
