`\text{~~Holi~~}`
`a. (x-1)^6=(x-1)^8`
`-> (x-1)^6-(x-1)^8=0`
`-> (x-1)^6.[1-(x-1)^2]=0`
`->`\(\left[ \begin{array}{l}(x-1)^6=0\\1-(x-1)^2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x-1=-1\\x-1=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x=0\\x=2\end{array} \right.\)
Vậy `x_1=0,x_2=0,x_3=2`
_______________________
Ở bình luận bn chủ tus cho em lm câu a thôi ạ.