Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{2}{{x + \sqrt x + 1}}\\
\Rightarrow {A^2} - 2A\\
= {\left( {\dfrac{2}{{x + \sqrt x + 1}}} \right)^2} - 2.\dfrac{2}{{x + \sqrt x + 1}}\\
= \dfrac{{{2^2}}}{{{{\left( {x + \sqrt x + 1} \right)}^2}}} - \dfrac{4}{{\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{4 - 4\left( {x + \sqrt x + 1} \right)}}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
= \dfrac{{ - 4x - 4\sqrt x }}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
= - 4.\dfrac{{x + \sqrt x }}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
Do:x \ge 0;x \ne 1\\
\Rightarrow x + \sqrt x > 0\\
\Rightarrow \dfrac{{x + \sqrt x }}{{{{\left( {x + \sqrt x + 1} \right)}^2}}} > 0\\
\Rightarrow - 4.\dfrac{{x + \sqrt x }}{{{{\left( {x + \sqrt x + 1} \right)}^2}}} < 0\\
\Rightarrow {A^2} - 2A < 0\\
\Rightarrow {A^2} < 2A
\end{array}$
