@py
Bài làm :
`(x-5)^(x+1) - (x-5)^(x+13) =0`
`⇔ (x-5)^(x+1) = (x-5)^(x+13)`
`⇔ (x-5)^(x+1) = (x-5)^(x+1) . (x-5)^(12)`
`⇔ (x-15)^(12) = 1`
`⇔` \(\left[ \begin{array}{l}x-5=1\\x-5=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=6\\x=4\end{array} \right.\)
Vậy `x ∈ { 6 ; 5 }`
