\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
<=> (x - 1)^2 = 1
<=> x - 1 = 1
<=> x = 2
`(x-1)^2 = (x-1)^4`
`=> (x-1)^4 - (x-1)^2 =0`
`=> (x-1)^2 [ (x-1)^2 - 1] =0`
`=> `\(\left[ \begin{array}{l}(x-1)^2=0\\(x-1)^2-1=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1\\x-1=1\\x-1=-1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\)
Vậy `x in { 1; 2;0}`