a 5 = a3 <=> a=1 hay a=(-1)
=> 2x -15= 1 hay 2x -15 = -1 => x= 8 hay x = 7
(2x-15)5 = (2x-15)3 => (2x-15)5 - (2x-15)3 = 0 => (2x-15)3.[(2x-15)2 - 1] = 0 => (2x-15)3 = 0 hoặc (2x-15)2 = 1
Nếu (2x-15)3 = 0 => 2x-15 = 0 => x = 15/2 \(\notin Z\)(loại)
Nếu (2x-15)2 = 1 => 2x - 15 = 1 hoặc 2x -15 = -1
2x-15 = 1 => x = 8
2x-15 = -1 => x=7
Vậy x = 7 , x= 8