Đáp án:
$\begin{array}{l}
a)DKxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
b)A = \left( {\frac{{2\sqrt x + x}}{{x\sqrt x - 1}} - \frac{1}{{\sqrt x - 1}}} \right):\left( {1 - \frac{{\sqrt x + 2}}{{x + \sqrt x + 1}}} \right)\\
= \frac{{2\sqrt x + x - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\frac{{x + \sqrt x + 1 - \sqrt x - 2}}{{x + \sqrt x + 1}}\\
= \frac{{2\sqrt x + x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{{x + \sqrt x + 1}}{{x - 1}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x - 1}}.\frac{1}{{x - 1}}\\
= \frac{1}{{x - 1}}\\
x = \sqrt {46 + 6\sqrt 5 } \left( {tmdk} \right)\\
= \sqrt {45 + 2.3\sqrt 5 + 1} \\
= \sqrt {{{\left( {3\sqrt 5 + 1} \right)}^2}} \\
= 3\sqrt 5 + 1\\
\Rightarrow A = \frac{1}{{x - 1}} = \frac{1}{{3\sqrt 5 + 1 - 1}} = \frac{{\sqrt 5 }}{{15}}
\end{array}$
