Question

tìm số tự nhiên n>1 sao cho 2n+1 chia hết cho n-1 

Answer 1

\(\Leftrightarrow2n-2+3⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;3;-3\right\}\)

mà n>1

nên \(n\in\left\{2;4\right\}\)

Answer 2

`2n + 1 vdots n - 1`.

`=> 2n - 2 + 3 vdots n  - 1`.

`=> 2(n-1) + 3 vdots n - 1`.

`=> 3 vdots n - 1 ( 2(n-1) vdots n - 1 )`.

`=> n - 1 in Ư(3)`

`=> n - 1 in {+-1, +-3}`

`=> n - 1 = 1 => n = 2.(tm)`

`=> n - 1 = -1 => n = 0(ktm)`

`=> n - 1 = 3 => n = 4(tm)`

`=> n - 1 = -3 => n = -2 (ktm)`

Vậy `n = 2, 4`.