Để\(A=\dfrac{3n+1}{n+1}\)có giá trị nguyên thì :
\(3n+1⋮n+1\)
\(\Leftrightarrow3n+3-2⋮n+1\)
\(\Leftrightarrow3\left(n+1\right)-2⋮n+1\)
Vì \(3\left(n+1\right)⋮n+1\)
\(\Rightarrow-2⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)
Xét bảng :
n+1 | 1 | -1 | 2 | -2 |
n | 0 | -2 | 1 | -3 |
Vậy....