\(\frac{3n+4}{2-n}=\frac{3n-6+10}{2-n}=\frac{-3\left(2-n\right)+10}{2-n}=-3+\frac{10}{2-n}\)
Để \(\frac{3n+4}{2-n}\) nguyên thì \(-3+\frac{10}{2-n}\) phải có GTN
\(\Rightarrow\frac{10}{2-n}\in Z\Rightarrow10⋮2-n\)
\(\Rightarrow2-n\inƯ\left(10\right)\Rightarrow2-n\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
\(\Rightarrow n\in\left\{1;3;0;4;-3;7;-8;12\right\}\)
Đặt A=\(\frac{3n+4}{2-n}=-3+\frac{10}{2-n}\)
Muốn A nguyên thì 2-n là Ư(10)=(-1;-2;-5;-10;1;2;5;10)