Đáp án: $\sqrt{2}$
Giải thích các bước giải:
ĐKXĐ: $x\ge\dfrac12$
Ta có:
$A=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}$
$\to A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}$
$\to A\sqrt{2}=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}$
$\to A\sqrt{2}=\sqrt{(\sqrt{2x-1}+1)^2}+\sqrt{(\sqrt{2x-1}-1)^2}$
$\to A\sqrt{2}=|\sqrt{2x-1}+1|+|\sqrt{2x-1}-1|$
$\to A\sqrt{2}=|\sqrt{2x-1}+1|+|1-\sqrt{2x-1}|$
$\to A\sqrt{2}\ge |\sqrt{2x-1}+1+1-\sqrt{2x-1}|$
$\to A\sqrt{2}\ge 2$
$\to A\ge \sqrt{2}$
Dấu = xảy ra khi: $(\sqrt{2x-1}+1)(1-\sqrt{2x-1})\ge 0$
$\to 1-\sqrt{2x-1}\ge 0$
$\to \sqrt{2x-1}\le 1$
$\to 0\le 2x-1\le 1$
$\to \dfrac12\le x\le 1$
