Đặt \(\sqrt{x^2-4x+16}=t>0\Rightarrow x^2-4x=t^2-16\)
Pt trở thành:
\(2\left(t^2-16\right)+t-4=0\)
\(\Leftrightarrow2t^2+t-36=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-\frac{9}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-4x+16}=4\)
\(\Leftrightarrow x^2-4x+16=16\)
\(\Leftrightarrow...\)