Đáp án:
$\begin{array}{l}
16)Dkxd:x > 0;x\# 1\\
A = \left( {\dfrac{x}{{\sqrt x - 1}} - \sqrt x } \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{{2 - x}}{{x - \sqrt x }}} \right)\\
= \dfrac{{x - \sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - x + \sqrt x }}{{\sqrt x - 1}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{x - 1 + \sqrt x + 2 - x}}\\
= \dfrac{{\sqrt x }}{1}.\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x + 1}}\\
17)\\
DKxd:x \ge 0;x\# 1\\
A = \left( {\dfrac{{2x + 1}}{{x\sqrt x - 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\left( {1 - \dfrac{{x - 2}}{{x + \sqrt x + 1}}} \right)\\
= \dfrac{{2x + 1 - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 - x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{x - \sqrt x }}{{\sqrt x - 1}}.\dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}
\end{array}$
