Đáp án:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
A = \left( {\dfrac{{\sqrt x + 1}}{{x - 2\sqrt x }} - \dfrac{1}{{\sqrt x - 2}}} \right).\left( {x - 3\sqrt x + 2} \right)\\
= \dfrac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)\\
= \dfrac{1}{{\sqrt x }}.\left( {\sqrt x - 1} \right)\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)A < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x }} - \dfrac{1}{2} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 2 - \sqrt x }}{{2\sqrt x }} < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{2\sqrt x }} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vay\,0 < x < 4\\
c)A = \dfrac{{\sqrt x - 1}}{{\sqrt x }} = 1 - \dfrac{1}{{\sqrt x }}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x }} \in Z\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vay\,x = 1
\end{array}$
