Đáp án:
$\begin{array}{l}
1)Đkxđ:x > 0;x \ne 4;x \ne 9\\
A = \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{4 - x}}} \right):\left( {\frac{{\sqrt x - 1}}{{x - 2\sqrt x }} - \frac{2}{{\sqrt x }}} \right)\\
= \left( {\frac{{4\sqrt x }}{{2 + \sqrt x }} + \frac{{8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}} \right):\left( {\frac{{\sqrt x - 1}}{{\left( {\sqrt x - 2} \right)\sqrt x }} - \frac{2}{{\sqrt x }}} \right)\\
= \frac{{4\sqrt x \left( {2 - \sqrt x } \right) + 8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}:\frac{{\sqrt x - 1 - 2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\sqrt x }}\\
= \frac{{8\sqrt x - 4x + 8x}}{{\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}.\frac{{\left( {\sqrt x - 2} \right)\sqrt x }}{{\sqrt x - 1 - 2\sqrt x + 4}}\\
= \frac{{8\sqrt x + 4x}}{{2 + \sqrt x }}.\frac{{ - \sqrt x }}{{3 - \sqrt x }}\\
= \frac{{4\sqrt x \left( {\sqrt x + 2} \right).\left( { - \sqrt x } \right)}}{{\left( {2 + \sqrt x } \right).\left( {3 - \sqrt x } \right)}}\\
= \frac{{4x}}{{\sqrt x - 3}}\\
2)A = - 2\\
\Rightarrow \frac{{4x}}{{\sqrt x - 3}} = - 2\\
\Rightarrow 4x = - 2\sqrt x + 6\\
\Rightarrow 2x + \sqrt x - 3 = 0\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {2\sqrt x + 3} \right) = 0\\
\Rightarrow \sqrt x = 1\left( {do\,\,\sqrt x > 0} \right)\\
\Rightarrow x = 1\left( {tmdk:x > 0;x \ne 4;x \ne 9} \right)
\end{array}$
Vậy x=1
