\(\left(x-1\right)^4=\left(1-x\right)^6\Leftrightarrow\left(x-1\right)^4=\left(x-1\right)^6\)
\(\Leftrightarrow\left(x-1\right)^4\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^4=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
a, (x-1)4=(1-x)6
⇒ (x-1)4=(x-1)6
⇒ (x-1)4 - (x-1)6 =0
⇒ (x-1)4 (1-(x-1)6)=0
⇒ \(\left[{}\begin{matrix}\left(x-1\right)^4=0\\1-\left(x-1\right)^6=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^6=1\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=1\\x-6=1\\x-6=-1\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=1\\x=7\\x=5\end{matrix}\right.\)
Vậy x ∈ \(\left\{1;7;5\right\}\)
