Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)Q = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right):\left( {\dfrac{{x + \sqrt x }}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{\sqrt x - x}}} \right)\\
= \left( {\dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right):\left( {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}} \right)\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)\\
= \dfrac{1}{{\sqrt x + 1}}:\dfrac{{x - 1}}{{\sqrt x }}\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{\sqrt x }}{{x - 1}}\\
= \dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}\\
b)Q = - 1\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}} = - 1\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( {x - 1} \right) = - \sqrt x \\
\Leftrightarrow x\sqrt x - \sqrt x + x - 1 = - \sqrt x \\
\Leftrightarrow x\sqrt x + x - 1 = 0\\
\Leftrightarrow \sqrt x = 0,75\\
\Leftrightarrow x = 0,563\left( {tmdk} \right)\\
Vậy\,x = 0,563
\end{array}$
