Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)C = \dfrac{1}{{\sqrt x - 1}} - \dfrac{3}{{x\sqrt x + 1}} + \dfrac{1}{{x - \sqrt x + 1}}\\
= \dfrac{{x - \sqrt x + 1 - 3 + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{x - 3}}{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
b)C < 1\\
\Leftrightarrow \dfrac{{x - 3}}{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}} < 1\\
\Leftrightarrow \dfrac{{x - 3 - \left( {x\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}} < 0\\
\Leftrightarrow \dfrac{{ - x\sqrt x + x - 2}}{{\left( {\sqrt x - 1} \right)}} < 0\\
\Leftrightarrow \dfrac{{x\sqrt x - x + 2}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{x\sqrt x + x - 2x - 2\sqrt x + 2\sqrt x + 2}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - 2\sqrt x + 2} \right)}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \sqrt x - 1 > 0\left( {do:\left\{ \begin{array}{l}
\sqrt x + 1 > 0\\
x - 2\sqrt x + 2 > 0
\end{array} \right.} \right)\\
\Leftrightarrow \sqrt x > 1\\
\Leftrightarrow x > 1\\
Vậy\,x > 1
\end{array}$
