Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 4\\
b)C = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{2 + 5\sqrt x }}{{x - 4}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
c)x = 4 - 2\sqrt 3 \left( {tmdk} \right)\\
= 3 - 2\sqrt 3 + 1\\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 - 1\\
\Leftrightarrow C = \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = \dfrac{{3\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1 + 2}}\\
= \dfrac{{3\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 + 1}}\\
= \dfrac{{3{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}\\
= \dfrac{{3.\left( {4 - 2\sqrt 3 } \right)}}{2}\\
= 3.\left( {2 - \sqrt 3 } \right)\\
= 6 - 3\sqrt 3
\end{array}$
