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