\(\dfrac{n-7}{n+3}=\dfrac{n+3-10}{n+3}=1-\dfrac{10}{n+3}\)
Để \(\dfrac{n-7}{n+3}\) nguyên thì \(n+3\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng sau:
\(n+3\) | \(-10\) | \(-5\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(5\) | \(10\) |
\(n\) | \(-13\) | \(-8\) | \(-5\) | \(-4\) | \(-2\) | \(-1\) | \(2\) | \(7\) |
Vậy để \(\dfrac{n-7}{n+3}\) nguyên thì \(n\in\left\{-13;-8;-5;-4;-2;-1;2;7\right\}\)