Bài làm :
Bài 5 :
c, `(x-1)^(x+2)=(x-1)^(x+6)` `;` `x \in ZZ`
`<=>(x-1)^(x+6)-(x-1)^(x+2)=0`
`<=>(x-1)^(x+2).(x-1)^4-(x-1)^(x+2)=0`
`<=>(x-1)^(x+2)[(x-1)^4-1]=0`
`<=>` \(\left[ \begin{array}{l}{x-1}^{x+2}=0\\(x-1)^4-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x-1=0\\(x-1)^4=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\(x-1)^4=(\pm1)^4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\x-1=1\\x-1=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1(tm)\\x=2(tm)\\x=0(tm)\end{array} \right.\)
Vậy `x \in {0;1;2}`
