\(B=\frac{3n+1}{n+1}=\frac{3n+3}{n+1}-\frac{2}{n+1}=3-\frac{2}{n+1}\)
B nguyên khi \(\frac{2}{n+1}\) nguyên <=> 2 chia hết cho n+1 <=>n+1 thuộc Ư(2)={-2;-1;1;2}
<=>n thuộc {-3;-2;0;1}
\(B=\frac{3n+1}{n+1}=\frac{3\left(n+1\right)-2}{n+1}=3-\frac{2}{n+1}\)
B nguyên <=> \(\frac{2}{n+1}\)nguyên
<=> \(2⋮n+1\)<=> \(n+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
n+1 | -2 | -1 | 1 | 2 |
n | -3 | -2 | 0 | 1 |