Giải thích các bước giải:
$A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\,\,\bigg{(}ĐK:\,x\ge \dfrac{1}{2}\bigg{)}$
$\sqrt 2A=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}$
$=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}$
$=\sqrt{(\sqrt{2x-1}+1)^2}-\sqrt{(\sqrt{2x-1}-1)^2}$
$=|\sqrt{2x-1}+1|-|\sqrt{2x-1}-1|$
TH1: $\sqrt{2x-1}\ge 1⇔x\ge 1$
$\sqrt 2A=\sqrt{2x-1}+1-\sqrt{2x-1}+1$
$=2$
$⇒A=\sqrt 2$
TH2: $\sqrt{2x-1}<1⇔x<1$
$\sqrt 2A=\sqrt{2x-1}+1+\sqrt{2x-1}-1$
$=2\sqrt{2x-1}$
$⇒A=\sqrt{4x-2}$.
