Vì \(n+2⋮n-1\)
\(\Rightarrow\left(n-1\right)+3⋮n-1\)
mà \(n-1⋮n-1\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\inƯ\left(3\right)\)
\(\Rightarrow n-1\in\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n\in\left\{2;0;4;-2\right\}\)
Vậy \(n\in\left\{0;\pm2;4\right\}.\)
Để \(\dfrac{n+2}{n-1}\) nhận giá trị nguyên thì :
\(n+2\text{ }⋮\text{ }n-1\)
\(\Rightarrow n-\left(1+3\right)\text{ }⋮\text{ }n-1\)
\(\Rightarrow n-1+3\text{ }⋮\text{ }n-1\)
\(\Rightarrow\left(n-1\right)+3\text{ }⋮\text{ }n-1\)
Mà \(n-1\text{ }⋮\text{ }n-1\)
\(\Rightarrow3\text{ }⋮\text{ }n-1\)
\(\Rightarrow\left(n-1\right)\inƯ_{\left(3\right)}\)
\(\Rightarrow\left(n-1\right)\in\left\{-1;1;-3;3\right\}\)
\(\Rightarrow n\in\left\{0;2;-2;4\right\}\)
Vậy \(\dfrac{n+2}{n-1}\) nhận giá trị nguyên khi \(n\in\left\{0;2;-2;4\right\}\)
