a, Ta có: n+12$\vdots$n
⇒n∈Ư(12)={±1;±2;±3;±4;±6;±12}
mà n∈N⇒n∈{1;2;3;4;6;12}
b, Ta có: n+13$\vdots$n-5
⇒(n-5)+18$\vdots$n-5
⇒n-5∈Ư(18)={1;2;3;6;9;18}
n-5=1⇒n=6 (tm)
n-5=2⇒n=7 (tm)
n-5=3⇒n=8 (tm)
n-5=6⇒n=11 (tm)
n-5=9⇒n=14 (tm)
n-5=18⇒n=23 (tm)
Vậy n∈{6;7;8;11;14;23}
c, Ta có: 6n+9$\vdots$4n-1
⇒2.(6n+9)$\vdots$4n-1
⇒12n+18$\vdots$4n-1
⇒3.(4n-1)+21$\vdots$4n-1
⇒4n-1∈Ư(21)={±1;±3;±7;±21}
4n-1=1⇒n=1/2 (loại)
4n-1=-1⇒n=0 (loại)
4n-1=3⇒n=1 (tm)
4n-1=-3⇒n=-1/2 (loại)
4n-1=7⇒n=2 (tm)
4n-1=-7⇒n=-3/2 (loại)
4n-1=21⇒n=11/2 (loại)
4n-1=-21⇒n=-5 9loaij)
Vậy n∈{1;2}