Question

Tìm tất cả số tự nhiên n để 3n + 13 chia hết n + 1

Answer 1

3n+13\(⋮\)n+1

3n+3+10\(⋮\)n+1

3(n+1)+10\(⋮\)3(n+1)

Vì 3(n+1)\(⋮\)n+1

Buộc 10\(⋮\)n+1=>n+1ϵƯ(10)={1;2;5;10}

Với n+1=1=>n=0

n+1=2=>n=1

n+1=5=>n=4

n+1=10=>n=9

Vậy nϵ{0;1;4;9}

Answer 2

Ta có:

\(3n+13⋮n+1\)

\(\Rightarrow\left(3n+3\right)+10⋮n+1\)

\(\Rightarrow3\left(n+1\right)+10⋮n+1\)

\(\Rightarrow10⋮n+1\)

\(\Rightarrow n+1\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)

+) \(n+1=1\Rightarrow n=0\)

+) \(n+1=2\Rightarrow n=1\)

+) \(n+1=5\Rightarrow n=4\)

+) \(n+1=10\Rightarrow n=9\)

Vậy n=0 ; n=1 ; n=4 ; n=9

Answer 3

3n+13n+1

3n+3+10n+1

3(n+1)+103(n+1)

Vì 3(n+1)n+1

Buộc 10n+1=>n+1ϵƯ(10)={1;2;5;10}

Với n+1=1=>n=0

n+1=2=>n=1

n+1=5=>n=4

n+1=10=>n=9

Vậy nϵ{0;1;4;9}