`a)` - Ta có : `n^2-7 vdots n+3`
mà `3(n+3) vdots n+3`
`=> n^2-7+3(n+3) vdots n+3`
`=> n^2-7+3n+9 vdots n+3`
`=> n(n+3)+(-7+9) vdots n+3`
`=> n(n+3)-2 vdots n+3`
mà `n(n+3) vdots n+3`
`=> -2 vdots n+3`
`=> n+3 in Ư(-2)={+-1;+-2}`
- Ta có bảng sau :
$\begin{array}{|c|c|c|c|} \hline n+3&-2&-1&1&2 \\\hline n&-5&-4&-2&-1 \\\hline \end{array}$
- Vậy `n in {-5;-4;-2;-1}`
- Vậy `n in {-5;-4;-2;-1}`
`b)` - Ta có : `n+3 vdots n^2-7`
`=> (n+3)(n-3) vdots n^2-7`
`=> n(n+3)-3(n+3) vdots n^2-7`
`=> n^2+3n-(3n+9) vdots n^2-7`
`=> n^2+(3n-3n)-9 vdots n^2-7`
`=> n^2-9 vdots n^2-7`
`=> n^2-7-2 vdots n^2-7`
mà `n^2-7 vdots n^2-7`
`=> -2 vdots n^2-7`
`=> n.n-7 in Ư(-2)={+-1;+-2}`
- Ta có bảng sau :
$\begin{array}{|c|c|c|c|} \hline n^2-7&-2&-1&1&2 \\\hline n^2&5&6&8&9 \\\hline n&\text{loại}&\text{loại}&\text{loại}&\pm3 \\\hline \end{array}$
- Vậy `n = +-3`
