Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
4 - x\# 0\\
\sqrt x \# 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x\# 4
\end{array} \right.\\
A = \left( {\dfrac{{4\sqrt x }}{{2 + \sqrt x }} + \dfrac{{8x}}{{4 - x}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \dfrac{2}{{\sqrt x }}} \right)\\
= \dfrac{{4\sqrt x .\left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}:\dfrac{{\sqrt x - 1 - 2\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{3 - \sqrt x }}\\
= \dfrac{{4\sqrt x \left( {\sqrt x + 2} \right)}}{{2 + \sqrt x }}.\dfrac{{ - \sqrt x }}{{3 - \sqrt x }}\\
= \dfrac{{4x}}{{\sqrt x - 3}}\\
b)A = - 2\\
\Leftrightarrow \dfrac{{4x}}{{\sqrt x - 3}} = - 2\\
\Leftrightarrow 2x = - \sqrt x + 3\\
\Leftrightarrow 2x + \sqrt x - 3 = 0\\
\Leftrightarrow 2x - 2\sqrt x + 3\sqrt x - 3 = 0\\
\Leftrightarrow \left( {\sqrt x - 1} \right)\left( {2\sqrt x + 3} \right) = 0\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
