Giải thích các bước giải:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x\sqrt x - 1 \ne 0\\
x + \sqrt x + 1 \ne 0\\
1 - \sqrt x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
C = \left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \left( {\dfrac{{x + 2}}{{{{\sqrt x }^3} - {1^3}}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \left( {\dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + \sqrt x .\left( {\sqrt x - 1} \right) - 1.\left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
b,\\
x = 4 \Rightarrow C = \dfrac{2}{{4 + \sqrt 4 + 1}} = \dfrac{2}{7}\\
c,\\
C = \dfrac{2}{{13}} \Leftrightarrow \dfrac{2}{{x + \sqrt x + 1}} = \dfrac{2}{{13}}\\
\Leftrightarrow x + \sqrt x + 1 = 13\\
\Leftrightarrow x + \sqrt x - 12 = 0\\
\Leftrightarrow \left( {\sqrt x + 4} \right)\left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 4 = 0\\
\sqrt x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 4\\
\sqrt x = 3
\end{array} \right.\\
\sqrt x \ge 0 \Rightarrow \sqrt x = 3 \Leftrightarrow x = 9\,\,\,\,\left( {t/m} \right)\\
d,\\
x \ge 0,x \ne 1 \Rightarrow x + \sqrt x + 1 \ge 1 > 0\\
\Rightarrow C = \dfrac{2}{{x + \sqrt x + 1}} > 0,\,\,\,\,\forall x > 0,x \ne 1
\end{array}\)
