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