a, Ta có: 2n-1$\vdots$n-5
⇒2.(n-5)+11$\vdots$n-5
⇒n-5∈Ư(11)={±1;±11}
n-5=1⇒n=6
n-5=-1⇒n=4
n-5=11⇒n=16
n-5=-11⇒n=-6
Vậy n∈{6;4;16;-6}
b, Ta có: n²+3n-13$\vdots$n+3
⇒n.(n+3)-13$\vdots$n+3
⇒n+3∈Ư(13)={±1;±13}
n+3=1⇒n=-2
n+3=-1⇒n=-4
n+3=13⇒n=10
n+3=-13⇒n=-16
Vậy n∈{-2;-4;10;-16}