Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 4\\
b)A = \dfrac{{\sqrt x }}{{\sqrt x + 2}} - \dfrac{{\sqrt x }}{{\sqrt x - 2}} + \dfrac{{2\sqrt x - 4}}{{x - 4}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x + 2} \right) + 2\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 2\sqrt x - x + 2\sqrt x + 2\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{2}{{\sqrt x + 2}}\\
c)x = 6 + 4\sqrt 2 \\
= 4 + 2.2.\sqrt 2 + 2\\
= {\left( {2 + \sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 + \sqrt 2 \\
A = \dfrac{2}{{\sqrt x + 2}} = \dfrac{2}{{2 + \sqrt 2 + 2}}\\
= \dfrac{2}{{4 + \sqrt 2 }}\\
= \dfrac{{2\left( {4 - \sqrt 2 } \right)}}{{16 - 2}}\\
= \dfrac{{4 - \sqrt 2 }}{7}\\
d)A = 2\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} = 2\\
\Leftrightarrow \sqrt x + 2 = 1\\
\Leftrightarrow \sqrt x = - 1\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
e)A = \dfrac{2}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 2\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tm} \right)\\
Vậy\,x = 0
\end{array}$
