Ta có: \(x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow A=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
Dấu " = " khi \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(MAX_A=\dfrac{4}{3}\) khi \(x=\dfrac{1}{4}\)