Đáp án:
$\begin{array}{l}
1)Q = \left( {\frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{{\sqrt x }}{{x - \sqrt x }}} \right):\frac{{\sqrt x + 1}}{{x - 1}}\left( {x > 0;x \ne 1} \right)\\
= \left( {\frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{1}{{\sqrt x - 1}}} \right).\frac{{x - 1}}{{\sqrt x + 1}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x - 1}}.\frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \sqrt x - 1\\
2)Q < 0\\
\Rightarrow \sqrt x - 1 < 0\\
\Rightarrow \sqrt x < 1\\
\Rightarrow x < 1\\
Vậy\,0 < x < 1
\end{array}$
