Đáp án:
$\begin{array}{l}
a)DKxd:x > 0;x \ne 1\\
A = \left( {\frac{{3 + \sqrt x }}{{x - 1}} + \frac{3}{{1 + \sqrt x }}} \right):\frac{4}{{x + \sqrt x }}\\
= \frac{{3 + \sqrt x + 3\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{4}\\
= \frac{{4\sqrt x }}{{\sqrt x - 1}}.\frac{{\sqrt x }}{4}\\
= \frac{x}{{\sqrt x - 1}}\\
b)x = \frac{9}{4}\left( {tmdk} \right)\\
\Rightarrow \sqrt x = \frac{3}{2}\\
\Rightarrow A = \frac{{\frac{9}{4}}}{{\frac{3}{2} - 1}} = \frac{{\frac{9}{4}}}{{\frac{1}{2}}} = \frac{9}{2}\\
c)A < 1\\
\Rightarrow \frac{x}{{\sqrt x - 1}} < 1\\
\Rightarrow \frac{{x - \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} < 0\\
\Rightarrow \frac{{x - \sqrt x + 1}}{{\sqrt x - 1}} < 0\\
\Rightarrow \sqrt x - 1 < 0\left( {do:x - \sqrt x + 1 > 0} \right)\\
\Rightarrow x < 1\\
Vậy\,0 < x < 1
\end{array}$
