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