Lời giải:
a, Để \(\dfrac{5}{n+1}\in Z\) thì \(5⋮\left(n+1\right)\) hay \(\left(n+1\right)\in U\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| \(n+1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(n\) | \(0\) | \(-2\) | \(4\) | \(-6\) |
Vậy, với \(n\in\left\{-6;-2;0;4\right\}\) thì \(\dfrac{5}{n+1}\in Z\).
b, Ta có: \(\dfrac{n-6}{n+1}=\dfrac{\left(n+1\right)-7}{n+1}=\dfrac{n+1}{n+1}-\dfrac{7}{n+1}=1+\dfrac{7}{n+1}\)
Để \(\dfrac{n-6}{n+1}\in Z\) thì \(\dfrac{7}{n+1}\in Z\Rightarrow7⋮\left(n+1\right)\) hay \(\left(n+1\right)\in U\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng sau:
| \(n+1\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(n\) | \(0\) | \(-2\) | \(6\) | \(-8\) |
Vậy, với \(n\in\left\{-8;-2;0;6\right\}\) thì \(\dfrac{n-6}{n+1}\in Z\).
c, Ta có: \(\dfrac{2n+7}{n+1}=\dfrac{\left(2n+2\right)+5}{n+1}=\dfrac{2n+2}{n+1}+\dfrac{5}{n+1}=2+\dfrac{5}{n+1}\)
Để \(\dfrac{2n+7}{n+1}\in Z\) thì \(\dfrac{5}{n+1}\in Z\Rightarrow5⋮\left(n+1\right)\) hay \(\left(n+1\right)\in U\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| \(n+1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(n\) | \(0\) | \(-2\) | \(4\) | \(-6\) |
Vậy, với \(x\in\left\{-6;-2;0;4\right\}\) thì \(\dfrac{2n+7}{n+1}\in Z\).
