Gọi ƯCLN (n+8,2n-5) = d(d∈N*)
=> $\left \{ {{n+8⋮d} \atop {2n-5 ⋮d}} \right.$
=> $\left \{ {{2n+16⋮d} \atop {2n-5⋮d}} \right.$
=> 2n + 16 - (2n-5) ⋮ d
=> 21⋮d
=> d∈{1,3,7}
Nếu d = 3
=> n+8 ⋮ 3
=> n+8 = 3k (k∈N*)
=> n = 3k - 8
=> 2n - 5 = 2(3k - 8) - 5 = 6k - 16 - 5 = 6k - 21 = 3(2k - 7) ⋮ 3
Vậy n khác 2k - 7 thì n+8/2n -5 tối giản.