Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
A = \left( {\dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}} \right):\left( {1 - \dfrac{{3 - \sqrt x }}{{\sqrt x + 1}}} \right)\\
= \left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x .\left( {\sqrt x + 1} \right)}}} \right]\\
:\dfrac{{\sqrt x + 1 - 3 + \sqrt x }}{{\sqrt x + 1}}\\
= \left( {\dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}} \right).\dfrac{{\sqrt x + 1}}{{2\sqrt x - 2}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}.\dfrac{{\sqrt x + 1}}{{2\sqrt x - 2}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b)x = 6 - 2\sqrt 5 = {\left( {\sqrt 5 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 - 1\\
\Rightarrow A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt 5 - 1 + 1}}{{\sqrt 5 - 1 - 1}} = \dfrac{{\sqrt 5 }}{{\sqrt 5 - 2}}\\
= \dfrac{{\sqrt 5 \left( {\sqrt 5 + 2} \right)}}{{5 - 4}}\\
= 5 + 2\sqrt 5 \\
c)x > 0;x \ne 1\\
A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}} = 1 + \dfrac{2}{{\sqrt x - 1}}\\
A \in Z\\
\Rightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Rightarrow \left( {\sqrt x - 1} \right) \in \left\{ { - 1;1;2} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;2;3} \right\}\\
\Rightarrow x \in \left\{ {0;4;9} \right\}\\
Do:x > 0;x \ne 1\\
\Rightarrow x \in \left\{ {4;9} \right\}
\end{array}$
