Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{\sqrt x + 1}}{{x + 4\sqrt x + 4}}:\left( {\dfrac{x}{{x + 2\sqrt x }} + \dfrac{x}{{\sqrt x + 2}}} \right)\\
= \dfrac{{\sqrt x + 1}}{{{{\sqrt x }^2} + 2.\sqrt x .2 + {2^2}}}:\left( {\dfrac{x}{{\sqrt x .\left( {\sqrt x + 2} \right)}} + \dfrac{x}{{\sqrt x + 2}}} \right)\\
= \dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}:\dfrac{{x + x.\sqrt x }}{{\sqrt x .\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}:\dfrac{{x.\left( {1 + \sqrt x } \right)}}{{\sqrt x .\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}.\dfrac{{\sqrt x .\left( {\sqrt x + 2} \right)}}{{x.\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{\sqrt x }}{{x.\left( {\sqrt x + 2} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
b,\\
A \ge \dfrac{1}{{3\sqrt x }}\\
\Leftrightarrow \dfrac{1}{{\sqrt x \left( {\sqrt x + 2} \right)}} \ge \dfrac{1}{{3\sqrt x }}\\
\Leftrightarrow \sqrt x \left( {\sqrt x + 2} \right) \le 3\sqrt x \\
\Leftrightarrow {\sqrt x ^2} + 2\sqrt x - 3\sqrt x \le 0\\
\Leftrightarrow {\sqrt x ^2} - \sqrt x \le 0\\
\Leftrightarrow \sqrt x \left( {\sqrt x - 1} \right) \le 0\\
\Leftrightarrow \sqrt x - 1 \le 0\,\,\,\,\,\,\left( {\sqrt x > 0,\,\,\,\forall x > 0} \right)\\
\Leftrightarrow \sqrt x \le 1\\
\Leftrightarrow 0 < x \le 1
\end{array}\)
