$(x-1)^{x+2}=(x-1)^{x+4}$
$(x-1)^{x+2}-(x-1)^{x+4}=0$
$(x-1)^{x+2}.[1-(x-1)²]=0$
⇒\(\left[ \begin{array}{l}(x-1)^{x+2}=0\\1-(x-1)²=0\end{array} \right.\)
TH1 :$ (x-1)^{x+2}=0$
⇒ $x-1=0$
⇒ $x=1$
TH2 : $1-(x-1)²=0$
⇒(x-1)²=1
⇒ \(\left[ \begin{array}{l}x-1=1\\x-1=-1\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy $x=1$ hoặc $x=2 $hoặc$ x=0$
