a) Ta có: \(A=\frac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\frac{x}{x+2\sqrt{x}}+\frac{x}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{x}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{1}{x+2\sqrt{x}}\)
b) Để \(A\ge\frac{1}{3\sqrt{x}}\) thì \(A-\frac{1}{3\sqrt{x}}\ge0\)
\(\Leftrightarrow\frac{1}{x+2\sqrt{x}}-\frac{1}{3\sqrt{x}}\ge0\)
\(\Leftrightarrow\frac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{1}{3\sqrt{x}}\ge0\)
\(\Leftrightarrow\frac{3}{3\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{3\sqrt{x}\left(\sqrt{x}+2\right)}\ge0\)
\(\Leftrightarrow\frac{1-\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)}\ge0\)
mà \(3\sqrt{x}\left(\sqrt{x}+2\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(1-\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}\le1\)
\(\Leftrightarrow x\le1\)
Kết hợp ĐKXĐ, ta được: \(0< x\le1\)
Vậy: Để \(A\ge\frac{1}{3\sqrt{x}}\) thì \(0< x\le1\)
