Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \ne 2\\
x \ne 4\\
\sqrt x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b)A = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}} \right).\dfrac{1}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\sqrt x .\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x .\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{3}{{\sqrt x + 2}}\\
c)A = \dfrac{3}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 3\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
