(x-3)^11=(x-3)^7
(x-3)^11-(x-3)^7=0
(x-3)^7[(x-3)^4-1)]=0
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^7=0\\\left(x-3\right)^4-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^4=1\end{cases}}\)\(\Rightarrow\)x=3; x=2; x=4
Vậy x=3 hoặc x=2 hoặc x=4
Ta có (x-3)^11 = (x-3)^7
<=> \(\hept{\begin{cases}x-3=0\\x-3=1\\x-3=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=4\\x=2\end{cases}}\)