Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x\# 1\\
A = \left( {\dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}} \right):\dfrac{{2\left( {x - 2\sqrt x + 1} \right)}}{{x - 1}}\\
= \left( {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right)\\
.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left( {\dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}} \right).\dfrac{{\sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}.\dfrac{{\sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b)A < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} < 0\\
\Leftrightarrow \sqrt x - 1 < 0\\
\Leftrightarrow \sqrt x < 1\\
\Leftrightarrow x < 1\\
Vậy\,0 < x < 1\\
c)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}}\\
= 1 + \dfrac{2}{{\sqrt x - 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in \left\{ {1;2} \right\}\left( {do:\sqrt x - 1 > - 1} \right)\\
\Leftrightarrow \sqrt x \in \left\{ {2;3} \right\}\\
\Leftrightarrow x \in \left\{ {4;9} \right\}\\
Vậy\,x \in \left\{ {4;9} \right\}
\end{array}$
