a, Ta có: 4n-5$\vdots$n
⇒n∈Ư(5)={±1;±5}
b, Ta có: -11$\vdots$n-1
⇒n-1∈Ư(11)={±1;±11}
n-1 1 -1 11 -11
n 2 0 12 -10
Vậy n∈{2;0;12;-10}
c, Ta có: 3n+2$\vdots$2n-1
⇒2(3n+2)$\vdots$2n-1
⇒6n+4$\vdots$2n-1
⇒3(2n-1)+7$\vdots$2n-1
⇒2n-1∈Ư(7)={±1;±7}
2n-1 1 -1 7 -7
2n 2 0 8 -6
n 1 0 4 -3
Vậy n∈{1;0;4;-3}