(a) \(A=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+2\sqrt{x}+1}\)
\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right]:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{1-x}{x}\)
Vậy: \(A=\dfrac{1-x}{x}\)
(b) \(A>\dfrac{1}{2}\Leftrightarrow\dfrac{1-x}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2\left(1-x\right)-x}{2x}>0\Rightarrow2\left(1-x\right)-x>0\) (do \(x>0\)).
\(\Leftrightarrow x< \dfrac{2}{3}\).
Vậy: \(A>\dfrac{1}{2}\Leftrightarrow0< x< \dfrac{2}{3}\)
