Đáp án:
$\begin{array}{l}
A = \frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{2}{{\sqrt x + 1}} - \frac{2}{{x - 1}} + 2\\
= \frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{2}{{\sqrt x + 1}} - \frac{2}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x \left( {\sqrt x + 1} \right) - 2\left( {\sqrt x - 1} \right) - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{x + \sqrt x - 2\sqrt x + 2 - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{x - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{\sqrt x }}{{\sqrt x + 1}}\\
Do:\sqrt x < \sqrt x + 1\left( {\forall x \ge 0;x \ne 1} \right)\\
\Rightarrow A = \frac{{\sqrt x }}{{\sqrt x + 1}} < 1
\end{array}$
Vậy A<1
