$\sqrt{2}A=\sqrt{2x-1-2\sqrt{2x-1}+1}-\sqrt{2x-1+\sqrt{2x-1}+1}$
$\sqrt{2}A=\sqrt{(\sqrt{2x-1}-1)^2}-\sqrt{(\sqrt{2x-1}+1)^2}$
$$=|\sqrt{2x-1}-1|-\sqrt{2x-1}-1$$
Đến đây xét dấu $\sqrt{2x-1}-1$
Nếu $\sqrt{2x-1}-1 \geq 0$
$\Rightarrow \sqrt{2}A=\sqrt{2x-1}-1-\sqrt{2x-1}-1=-2$
$\Rightarrow A=-\sqrt{2}$
Nếu $\sqrt{2x-1}-1<0$
$\Rightarrow \sqrt{2}A=1-\sqrt{2x-1}-\sqrt{2x-1}-1=-2\sqrt{2x-1}$
$\Rightarrow A=-\sqrt{2}\sqrt{2x-1}$