Giúp bài này ạ . . . . . .

$\begin{array}{l}
1)A = \dfrac{{\sqrt x  + 1}}{{x + 4\sqrt x  + 4}}:\left( {\dfrac{x}{{x + 2\sqrt x }} + \dfrac{x}{{\sqrt x  + 2}}} \right)\\
A = \dfrac{{\sqrt x  + 1}}{{{{\left( {\sqrt x  + 2} \right)}^2}}}:\left[ {\dfrac{x}{{\sqrt x \left( {\sqrt x  + 2} \right)}} + \dfrac{x}{{\sqrt x  + 2}}} \right]\\
A = \dfrac{{\sqrt x  + 1}}{{{{\left( {\sqrt x  + 2} \right)}^2}}}:\dfrac{{x + x\sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
A = \dfrac{{\sqrt x  + 1}}{{{{\left( {\sqrt x  + 2} \right)}^2}}}.\dfrac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{x\left( {\sqrt x  + 1} \right)}} = \dfrac{1}{{\sqrt x \left( {\sqrt x  + 2} \right)}}\\
2)A \ge \dfrac{1}{{3\sqrt x }} \Leftrightarrow \dfrac{1}{{\sqrt x \left( {\sqrt x  + 2} \right)}} \ge \dfrac{1}{{3\sqrt x }}\\
 \Leftrightarrow \dfrac{1}{{\sqrt x  + 2}} \ge \dfrac{1}{3} \Leftrightarrow \sqrt x  + 2 \le 3 \Leftrightarrow \sqrt x  \le 1\\
 \Rightarrow x < 1\\
\text{Kết hợp điều kiện} :x > 0 \Rightarrow 0 < x < 1
\end{array}$