\(\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
Vậy GTLN của biểu thức là \(\dfrac{4}{3}\) . Dấu \("="\) xảy ra khi \(x=\dfrac{1}{4}\)
x\(x\ge0\)
\(x-\sqrt{x}+1=\sqrt{x}^2-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{x-\sqrt{x}+1}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(\Rightarrow\) biểu thức đạt GTLN bằng \(\dfrac{4}{3}\) khi \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
ĐK: \(x\ge0\)
Ta có: \(\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{\left(\sqrt{x}^2-2.\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
Vậy, GTLN là \(\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{4}\)
