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