Bài 2:
a. \(A\in Z\Leftrightarrow\dfrac{n+1}{n-3}\in Z\Leftrightarrow n+1⋮n-3\)
Ta có: \(n+1=\left(n-3\right)+3+1=\left(n-3\right)+4\)
Để \(n+1⋮n-3\) thì \(4⋮n-3\)
\(\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau:
n-3 |
-4 |
-2 | -1 | 1 | 2 | 4 |
n | -1 | 1 | 2 | 4 | 5 | 7 |
Vậy \(n\in\left\{-1;1;2;4;5;7\right\}\)