Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x\sqrt x - 1 \ne 0\\
x + \sqrt x + 1 \ne 0\\
1 - \sqrt x \ne 0\\
\dfrac{{\sqrt x - 1}}{2} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
A = \left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \left( {\dfrac{{x + 2}}{{{{\sqrt x }^3} - {1^3}}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \left( {\dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{\left( {x + 2} \right) + \sqrt x .\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right).1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right).\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right).\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
b,\\
x \ge 0 \Rightarrow x + \sqrt x + 1 \ge 1 > 0 \Rightarrow \dfrac{2}{{x + \sqrt x + 1}} > 0\\
\Rightarrow A > 0\\
c,\\
x \ge 0 \Rightarrow x + \sqrt x + 1 \ge 1 \Rightarrow \dfrac{2}{{x + \sqrt x + 1}} \le \dfrac{2}{1} = 2\\
\Rightarrow A \le 2,\,\,\,\,\forall x \ge 0,x \ne 1\\
\Rightarrow {A_{\max }} = 2 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0
\end{array}\)
