Đáp án:
$\min(x - \sqrt x + 1) = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{4}$
Giải thích các bước giải:
$x - \sqrt x + 1 \qquad (x \geq 0)$
$= (\sqrt x)^2 - 2.\dfrac{1}{2}.\sqrt x + \dfrac{1}{4} + \dfrac{3}{4}$
$= \left(\sqrt x - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}$
Ta có:
$\left(\sqrt x - \dfrac{1}{2}\right)^2 \geq,\,\forall x $
$\to \left(\sqrt x - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}$
Dấu = xảy ra $\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0\Leftrightarrow x = \dfrac{1}{4}$
Vậy $\min(x - \sqrt x + 1) = \dfrac{3}{4} \Leftrightarrow x = \dfrac{1}{4}$
