Ta có: \(n+2⋮n-1\)
\(\Rightarrow n-1+3⋮n-1\)
\(\Rightarrow n-1⋮n-1\)
\(\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\inƯ_{\left(3\right)}=-3;-1;1;3\)
\(\Rightarrow\left[\begin{matrix}n-1=-3\\n-1=-1\\n-1=1\\n-1=3\end{matrix}\right.\Rightarrow\left[\begin{matrix}n=-2\\n=0\\n=2\\n=4\end{matrix}\right.\)
Vậy: \(n=-2;0;2;4\)
Để (n+2) chia hết cho (n-1)
thì (n-1)+3 chia hết cho (n-1)
mà n-1 chia hết cho n-1
nên 3 chia hết cho n-1 hay n-1 thuộc Ư(3)={-3;-1;1;3}
n-1 | -3 | -1 | 1 | 3 |
n | -2 | 0 | 2 | 4 |