Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
C = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 - x} \right)}^2}}}{2}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{x - \sqrt x - 2 - \left( {x + \sqrt x - 2} \right)}}{1}.\dfrac{{\sqrt x - 1}}{2}\\
= \left( { - 2\sqrt x } \right).\dfrac{{\sqrt x - 1}}{2}\\
= \sqrt x - x\\
b)C > 0\\
\Rightarrow \sqrt x - x > 0\\
\Rightarrow \sqrt x \left( {1 - \sqrt x } \right) > 0\\
\Rightarrow \left\{ \begin{array}{l}
1 - \sqrt x > 0\\
x \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt x < 1\\
x \ne 0
\end{array} \right.\\
\Rightarrow x < 1;x \ne 0\\
c)C = \sqrt x - x\\
= - \left( {x - \sqrt x } \right)\\
= - \left( {x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= - {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\Rightarrow GTLN:C = \dfrac{1}{4}\\
Khi:x = \dfrac{1}{4}
\end{array}$
