ĐK: x ≥ 0
Ta có:
\(x-\sqrt{x}+2=x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{7}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Dấu " = " xảy ra ⇔ \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy Min = 7/4 ⇔ x = 1/4