Đáp án + Giải thích các bước giải:
`(x-7)^{x+1}-(x-7)^{x+11}=0`
`=>(x-7)^{x+11}-(x-7)^{x+1}=0`
`=>(x-7)^{x+1}[(x-7)^{10}-1]=0`
`=>` \(\left[ \begin{array}{l}(x-7)^{x+1}=0\\(x-7)^{10}-1=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x-7=0\\(x-7)^{10}=1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=7\\x-7=±1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=7\\x=8\\x=6\end{array} \right.\)
Vậy `x∈{6;7;8}`
