Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
b)A = \left( {1 + \frac{{x + \sqrt x }}{{\sqrt x + 1}}} \right).\left( {1 - \frac{{x - \sqrt x }}{{\sqrt x - 1}}} \right)\\
= \left( {1 + \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right).\left( {1 - \frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}} \right)\\
= \left( {1 + \sqrt x } \right).\left( {1 - \sqrt x } \right)\\
= 1 - x\\
c)x \ge 0;x \ne 1\\
A = 1 - x\\
Do:x \ge 0\\
\Rightarrow - x \le 0\\
\Rightarrow A = 1 - x \le 1\\
\Rightarrow GTLN:A = 1 \Leftrightarrow x = 0
\end{array}$
