Để \(4n+11⋮n+2\Leftrightarrow\frac{4n+11}{n+2}\in Z\Leftrightarrow\frac{4n+8+3}{n+2}\in Z\Leftrightarrow4+\frac{3}{n+2}\in Z\)
\(\Leftrightarrow\frac{3}{n+2}\in Z\) (vì \(4\in Z\))
\(\Leftrightarrow3⋮n+2\)
\(\Leftrightarrow n+2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow n\in\left\{-1;-3;1;-5\right\}\)
Vậy để \(4n+11⋮n+2\Leftrightarrow x\in\left\{-1;-3;1;-5\right\}\)
4n+11⋮n+2
4n+8+3⋮n+2
4n+4.2+3⋮n+2
4(n+2)+3⋮n+2
Vì 4(n+2)+3⋮n+2
n+2⋮n+2 => 4(n+2)⋮n+2
=>3⋮n+2
=>n+2∈Ư(3)
=>n+2=1; -1; 3; -3
Ta có bảng sau:
| n+2 | 1 | -1 | 3 | -3 |
| n | -1 | -3 | 1 | -5 |
Vậy n=-1; -3; 1; -5.
