Question

tìm số tự nhiên n để n + 9 chia hết cho n + 3

Answer 1

Vì : \(n+9⋮n+3\)

Mà : \(n+3⋮n+3\)

\(\Rightarrow\left(n+9\right)-\left(n+3\right)⋮n+3\)

\(\Rightarrow n+9-n-3⋮n+3\)

\(\Rightarrow6⋮n+3\)

\(\Rightarrow n+3\inƯ\left(6\right)\)

Mà : \(Ư\left(6\right)=\left\{1;2;3;6\right\}\) ; \(n+3\ge3\)

\(\Rightarrow n+3\in\left\{3;6\right\}\)

+) \(n+3=3\Rightarrow n=3-3\Rightarrow n=0\)

+) \(n+3=6\Rightarrow n=6-3\Rightarrow n=3\)

Vậy : \(n\in\left\{0;3\right\}\)

Answer 2

n+9\(⋮\)n+3

n+3-12\(⋮\)n+3

Vì n+3\(⋮\)n+3

Buộc 12\(⋮\)n+3=>n+3ϵƯ(12)={1;2;3;4;6;12}

ta có bảng sau :

n+3	1	2	3	4	6	12
n	-2	-1	0	1	3	9

Vậy n ϵ {0;1;3;9}