Giải thích các bước giải:
`(x-1)^(x+2)=(x - 1)^(x+6)`
`=>(x - 1)^(x+6)-(x-1)^(x+2)=0`
`=>(x - 1)^(x+2).(x - 1)^4-1.(x-1)^(x+2)=0`
`=>(x - 1)^(x+2).[(x - 1)^4-1]=0`
`=>`\(\left[ \begin{array}{l}(x - 1)^{x+2}=0\\(x - 1)^4-1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x-1=0\\\left[ \begin{array}{l}x-1=1\\x-1=-1\end{array} \right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\)
Vậy `x in{0;1;2}.`
