Đáp án:
$\begin{array}{l}
A = \frac{{\sqrt x + 1}}{{\sqrt x - 1}} + \frac{{\sqrt x - 1}}{{\sqrt x + 1}} + \frac{{3\sqrt x + 1}}{{1 - x}}\left( {x \ge 0;x \ne 1} \right)\\
= \frac{{\sqrt x + 1}}{{\sqrt x - 1}} + \frac{{\sqrt x - 1}}{{\sqrt x + 1}} - \frac{{3\sqrt x + 1}}{{x - 1}}\\
= \frac{{{{\left( {\sqrt x + 1} \right)}^2} + {{\left( {\sqrt x - 1} \right)}^2} - 3\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{2x + 2 - 3\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{2x - 3\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{2x - 2\sqrt x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {2\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{2\sqrt x - 1}}{{\sqrt x + 1}}\\
2)\\
A = \frac{{2\sqrt x - 1}}{{\sqrt x + 1}} = \frac{{2\left( {\sqrt x + 1} \right) - 2 - 1}}{{\sqrt x + 1}}\\
= 2 - \frac{3}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\forall x \ge 0;x \ne 1\\
\Rightarrow \sqrt x + 1 \ge 1\\
\Rightarrow \frac{3}{{\sqrt x + 1}} \le 3\\
\Rightarrow - \frac{3}{{\sqrt x + 1}} \ge - 3\\
\Rightarrow 2 - \frac{3}{{\sqrt x + 1}} \ge 2 - 3\\
\Rightarrow A \ge - 1\\
\Rightarrow GTNN\,A = - 1 \Leftrightarrow x = 0
\end{array}$
