a) điều kiện xác định : \(x\ge0;x\ne1\)
ta có : \(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(1-x\right)^2}{2}\) \(\Leftrightarrow A=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)\left(x-1\right)}{2}\)
\(\Leftrightarrow A=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) để \(A>0\Leftrightarrow-x+\sqrt{x}>0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}\ne0\\1-\sqrt{x}>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne0\\1>x\end{matrix}\right.\) \(\Leftrightarrow0< x< 1\)
c) ta có : \(A=-x+\sqrt{x}=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow A_{max}=\dfrac{1}{4}\) dấu "=" xảy ra khi \(x=\dfrac{1}{4}\)