Đáp án:
$\sqrt{2x + \sqrt{4x - 1}} + \sqrt{2x - \sqrt{4x - 1}} = \sqrt2 \Leftrightarrow \dfrac14 \leq x < \dfrac12$
Giải thích các bước giải:
Đặt $A = \sqrt{2x + \sqrt{4x - 1}} + \sqrt{2x - \sqrt{4x - 1}}\qquad \left(x \geq \dfrac{1}{4}\right)$
$\to A\sqrt2 = \sqrt{4x + 2\sqrt{4x - 1}} + \sqrt{4x - 2\sqrt{4x - 1}}$
$\to A\sqrt2 = \sqrt{4x - 1 + 2\sqrt{4x - 1} + 1} + \sqrt{4x - 1 - 2\sqrt{4x - 1} + 1}$
$\to A\sqrt2 = \sqrt{(\sqrt{4x - 1} + 1)^2} + \sqrt{(\sqrt{4x - 1} - 1)^2}$
$\to A\sqrt2 = \sqrt{4x - 1} + 1 + |\sqrt{4x - 1} - 1|$
$\to A\sqrt2 = \left[\begin{array}{l}\sqrt{4x - 1} + 1 + \sqrt{4x - 1} - 1\\\sqrt{4x - 1} + 1 + 1 - \sqrt{4x - 1}\end{array}\right.$
$\to A\sqrt2 = \left[\begin{array}{l}2\sqrt{4x - 1} \qquad \left(x \geq \dfrac12\right)\\2\qquad \qquad \left(\dfrac14 \leq x < \dfrac12\right)\end{array}\right.$
$\to A = \sqrt2 \Leftrightarrow \dfrac14 \leq x < \dfrac12$
