Ta có :
$(x-1)^{x+2}=(x-1)^{x+6}$
$⇔(x-1)^{x+2}.[(x-1)^4 - 1 ] = 0$
$⇔ \left[ \begin{array}{l}(x-1)^{x+2} = 0\\(x-1)^4-1=0\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=1\\(x-1)^4=1\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=1\\x-1=1\\x-1=-1\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.$
Vậy : $x ∈ $ { $1,2,0$}
