ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a, \(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b. \(A>0\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0\Rightarrow\sqrt{x}-1< 0\Rightarrow0\le x< 1\)
c. \(A=-\left(x-\sqrt{x}\right)=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\Rightarrow A\le\frac{1}{4}\)
Vậy \(MaxA=\frac{1}{4}\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
