Đáp án:
$\begin{array}{l}
A = \left( {\frac{{\sqrt x - 1}}{{\sqrt x - 2}} + \frac{2}{{x - 4}}} \right):\frac{{\sqrt x }}{{\sqrt x + 2}}\left( {x > 0;x \ne 4} \right)\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 2}}{{\sqrt x }}\\
= \frac{{x + \sqrt x - 2 + 2}}{{\sqrt x - 2}}.\frac{1}{{\sqrt x }}\\
= \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x + 1}}{{\sqrt x - 2}}\\
b)x > 0;x \ne 4\\
A < 1\\
\Rightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 2}} < 1\\
\Rightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 2}} - 1 < 0\\
\Rightarrow \frac{{\sqrt x + 1 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\\
\Rightarrow \frac{3}{{\sqrt x - 2}} < 0\\
\Rightarrow \sqrt x - 2 < 0\\
\Rightarrow \sqrt x < 2\\
\Rightarrow 0 < x < 4\\
Vậy\,0 < x < 4
\end{array}$
