tìm n thuộc Z a)2n-1 chia hết cho n-5 b)n^2+3n-13 chia hết cho n +3

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}