\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=> (x-7)x . (x-7) - (x-7)x . (x-7)11 = 0
=> \(\left(x-7\right)^x.\left[\left(x-7\right)-\left(x-7\right)^{11}\right]=0\)
=> [(x-7) - (x-7)11 ] = 0
=> \(\left\{\left(x-7\right).\left[1-\left(x-7\right)^{10}\right]\right\}=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\1-\left(x-7\right)^{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{10}=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{10}=\left(-1\right)^{10}=1^{10}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=7\\x-7=1\\x-7=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=8\\x=6\end{cases}}\)
Vậy x thuộc { 6,7,8}
