\(A=x-\sqrt{x}\)
\(+,0< x< 1\Rightarrow\sqrt{x}>x\Rightarrow x-\sqrt{x}< 0\Rightarrow A< 0\Rightarrow A< \left|A\right|\)
\(+,x\ge1\Rightarrow x\ge1\Rightarrow x\ge\sqrt{x}\Rightarrow x-\sqrt{x}\ge0\Rightarrow A\ge0\Rightarrow A=\left|A\right|\)
\(b,A=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}+\frac{1}{4}=2+\frac{1}{4}=\frac{9}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=\left(\pm\frac{3}{2}\right)^2\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=4\) \(c,A=x-\sqrt{x}\Rightarrow A=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge0-\frac{1}{4}=\frac{-1}{4}\Rightarrow A_{min}=\frac{-1}{4}.\text{Dâu "=" xay ra khi:}\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
