Question

Cho \(A=\dfrac{5n+1}{n+1}\left(n\ne-1\right)\). Tìm n ϵ N để A nguyên

Answer 1

$A=\frac{5n+1}{n+1}=\frac{5(n+1)-4}{n+1}=5-\frac{4}{n+1}\in \mathbb{Z}$

$\Leftrightarrow n+1\in Ư(4)=\left\{-4;-2;-1;1;2;4\right\}$

Mà $n\in\mathbb{N}$

$\Rightarrow n\in\left\{0;1;3\right\}$

Answer 2

\(A=\dfrac{5n+1}{n+1}=\dfrac{5\left(n+1\right)-4}{n+1}=\dfrac{5\left(n+1\right)}{n+1}-\dfrac{4}{n+1}=5-\dfrac{4}{n+1}\).ĐK:n≠-1

để \(Anguy\text{ê}n.th\text{ì}4⋮(n+1)\\ \Rightarrow n+1\in\text{Ư}\left(4\right)=\left\{1;2;4\right\}\)

ta có bảng sau :

n+1	1	2	4
n	0	1	3

vậy....