Ta có: \(\dfrac{n+5}{n+1}=\dfrac{\left(n+1\right)+4}{x+1}=1+\dfrac{4}{x+1}\)
Để \(\dfrac{n+5}{n+1}\) nguyên thì \(x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(x+1=-1=>x=-2\\ x+1=-2=>x=-3\\ x+1=-4=>x=-5\\ x+1=1=>x=0\\ x+1=2=>x=1\\ x+1=4=>x=3\)
Vậy: \(n\in\left\{-5;-3;-2;0;1;3\right\}\)
\(\dfrac{n+5}{n+1}=\dfrac{\left(n+1\right)+4}{n+1}=1+\dfrac{4}{n+1}\)
\(\Rightarrow n+1\in\text{Ư}\left(4\right)=\left\{1;2;4\right\}\)
\(\Rightarrow n=\left\{0;1;3\right\}\)
Vậy : .........
Điều kiện \(n+1\ne0\Rightarrow n\ne-1\)
Ta có: \(\dfrac{n+5}{n+1}=\dfrac{n+1+4}{n+1}=\dfrac{n+1}{n+1}+\dfrac{4}{n+1}=1+\dfrac{4}{n+1}\)
Do 1 nguyên \(\Rightarrow\) để \(\dfrac{n+5}{n+1}\) nguyên thì \(\dfrac{4}{n+1}\) nguyên \(\Rightarrow4⋮n+1\Rightarrow\) n+1 là ước của 4.
Mà \(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\Rightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng giá trị tương ứng của n:
| n+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 0 (TM) | -2 (TM) | 1 (TM) | -3 (TM) | 3 (TM) | -5 (TM) |
Vậy \(n\in\left\{0;-2;1;-3;3;-5\right\}\)
Để phân số \(\dfrac{n+5}{n+1}\)nguyên thì:
n+5\(⋮\)n-1
n-1+6\(⋮\)n-1
Vì n-1\(⋮\)n-1 nên 6\(⋮\)n-1
=>n-1 là Ư(6)
Ư(6)={1;-1;2;-2;3;-3;6;-6}
n={2;0;3;-1;4;-2;7;-5}
mk nhầm đề nhé
Để\(\dfrac{n+5}{n+1}\in\)Z thì n+5 \(⋮\)n+1
n+4+1\(⋮\)n+1
Vì n+1\(⋮\)n+1 nên 4\(⋮\)n+1
=> n+1 là Ư(4)
Ư(4)={-1;1;-2;2;-4;4}
n={0;2;-1;3;-3;5}
