Đáp án:
$x \leq \dfrac{1}{4}$
Giải thích các bước giải:
$\begin{array}{l}\sqrt{x^2 - \dfrac{1}{2}x + \dfrac{1}{16}} = \dfrac{1}{4} - x\\ \Leftrightarrow \sqrt{x^2 - 2.\dfrac{1}{4}x + \dfrac{1}{4^2}} = \dfrac{1}{4} - x\\ \Leftrightarrow \sqrt{\left(x - \dfrac{1}{4}\right)^2} = \dfrac{1}{4} - x\\ \Leftrightarrow \left|x - \dfrac{1}{4}\right| = \dfrac{1}{4} - x\\ \Leftrightarrow \left[\begin{array}{l}x - \dfrac{1}{4} = \dfrac{1}{4} - x \qquad \left(x \geq \dfrac{1}{4}\right)\\\dfrac{1}{4} - x = \dfrac{1}{4} - x \qquad \left(x < \dfrac{1}{4}\right)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{1}{4} &\left(x \geq \dfrac{1}{4}\right)\\\dfrac{1}{4} - x = \dfrac{1}{4} - x \quad \text{(hiển nhiên)}&\left(x < \dfrac{1}{4}\right)\end{array}\right.\\ \text{Vậy $x \leq \dfrac{1}{4}$} \end{array}$
