(x-1)x+2 = (x-1)x+6
=> (x-1)x+6-(x-1)x+2=0
=> (x-1)x+2[(x-1)4-1]=0
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)
\(\Rightarrow\)x=1 hoặc \(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy \(x\in\left\{1;2;0\right\}\)