a ) \(A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}.\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
b ) Ta có : \(A\ge\dfrac{1}{3\sqrt{x}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\ge\dfrac{1}{3\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}\ge x+2\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}\le0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)\le0\)
\(\Leftrightarrow\sqrt{x}-1\le0\Leftrightarrow x\le1\) và \(x>0\)
Vậy để \(A\ge\dfrac{1}{3\sqrt{x}}\) thì các giá trị của x tương ứng phải \(0< x\le1\)
