\(3n-13⋮\left(n+2\right)\)
\(\Rightarrow3\left(n+2\right)-19⋮\left(n+2\right)\)
\(\Rightarrow19⋮\left(n+2\right)\Rightarrow\left(n+2\right)\inƯ\left(19\right)=\left\{\pm1;\pm19\right\}\)
\(\Rightarrow n\in\left\{-1;-3;17;-21\right\}\)
3n-13 chia hết cho n+2
Mà n+2 chia hết cho n+2
Nên 3(n+2) chia hết cho n+2
3n+6 chia hết cho n+2
=> (3n-13)-(3n+6) chia hết cho n+2
=> -19 chia hết cho n+2
=> n+2 € Ư(-19)
n+2 € {1;-1;19;-19}
Vậy n € {-1;-3;17;-21}
\(3n-13⋮\left(n+2\right)\)
\(\Rightarrow3\left(n+2\right)-19⋮\left(n+2\right)\)
\(\Rightarrow19⋮\left(n+2\right)\Rightarrow\left(n+2\right)\in\text{Ư}\left(19\right)=\left\{\pm1;\pm19\right\}\)
\(\Rightarrow n\in\left\{-1;-3;17;-21\right\}\)
\(3n-13⋮n+2\)
\(\Rightarrow3\left(n+2\right)-19⋮n-2\)
\(\Rightarrow19⋮n-2\)
\(\Rightarrow n-2\inƯ\left(19\right)=\left\{\pm1;\pm19\right\}\)
Xét bảng
| n-2 | 1 | 19 | -1 | -19 |
| n | 3 | 21 | 1 | -17 |
Vậy............................