Giải thích các bước giải:
\(\begin{array}{l}
a.A = \frac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}:\left[ {\frac{{x + x\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]\\
= \frac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}.\frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{x\left( {\sqrt x + 1} \right)}}\\
= \frac{1}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
b.A \ge \frac{1}{{3\sqrt x }}\\
\to \frac{1}{{\sqrt x \left( {\sqrt x + 2} \right)}} \ge \frac{1}{{3\sqrt x }}\\
\to \frac{{3 - \sqrt x - 2}}{{3\sqrt x \left( {\sqrt x + 2} \right)}} \ge 0\\
\to \frac{{1 - \sqrt x }}{{3\sqrt x \left( {\sqrt x + 2} \right)}} \ge 0\\
\to 1 - \sqrt x \ge 0\left( {do:3\sqrt x \left( {\sqrt x + 2} \right) > 0\forall x > 0} \right)\\
\to 1 \ge \sqrt x \\
\to 0 < x \le 1
\end{array}\)
