Đáp án:
`x in {1;2;3}`
Giải thích các bước giải:
`(x-2)^2=(x-2)^4`
`(x-2)^4-(x-2)^2=0`
`(x-2)^2.(x-2)^2-(x-2)^2. 1=0`
`(x-2)^2[(x-2)^2-1]=0`
`=>`\(\left[ \begin{array}{l}(x-2)^2=0\\(x-2)^2-1=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}(x-2)^2=0\\(x-2)^2=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}|x-2|=0\\|x-2|=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x-2=0\\x-2=1\\x-2=-1\end{array}\right.\)
`=>`\(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array}\right.\)
Vậy `x in{1;2;3}`
