$Đk:0\leq x\leq 2$.Đặt$:(x-1)^2=t(t\geq 0)$PT trở thành$:\sqrt{2+2\sqrt{t}}=2t^2(2t^2-1)\Leftrightarrow 2+2\sqrt{t}=(4t^4-2t^2)^2\Leftrightarrow 2(\sqrt{t}-1)(8t^{15/2}+8t^{13/2}+2t^{7/2}+2t^{5/2}+2t^{3/2}+8t^7+8t^6+2t^3+2t^2+2t+2\sqrt{t}+1)=0$
Do cái ngoặc thứ $2\geq1\Rightarrow t=1$
$\Rightarrow (x-1)^2=1\Leftrightarrow x(x-2)=0\Leftrightarrow x=0hoặc x=2$