Question

Tìm n sao cho:
n+3 chia hết cho n+1

Answer 1

Giải:
Ta có: \(n+3⋮n+1\)

\(\Rightarrow\left(n+1\right)+2⋮n+1\)

\(\Rightarrow2⋮n+1\)

\(\Rightarrow n+1\in\left\{1;-1;2;-2\right\}\)

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

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

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

+) \(n+1=-2\Rightarrow n=-3\)

Vậy \(n\in\left\{0;-2;1;-3\right\}\)

Answer 2

n+3 chia hết cho n+1

<=> (n+1)+2 chia hết cho n+1

=> n+1 thuộc Ư(2)

Ta có :

Ư(2) = {1;2)

Nếu n+1=1

=> n = 0

Nếu n+1=2

=> n = 1

Vậy n thuộc {0;1}

=>

Answer 3

Để n+3 chia hết cho n+1 thì:

n+1+2\(⋮\)n+1

=> 2\(⋮\)n+1( vì n+1\(⋮\)n+1)

=> n+1\(_{\in}\)Ư(2)={1;2}

* Xét n+1=1=>n=0

*Xét n+1=2=>n=1

Vậy n\(_{\in}\){0;1}

Answer 4

n + 3 = (n + 1) + 2

Mà n + 1 chia hết cho n+1

Suy ra : 2 chia hết cho n + 1

Suy ra: n + 1 là Ư(2) = (1;2)

Suy ra : n thuộc (0;1)

Xong rồi đó bn

Answer 5

Ta có: \(n+3⋮n+1\)

\(\Rightarrow\left[\left(n+1\right)+2\right]⋮\left(n+1\right)\)

\(\Rightarrow2⋮n+1\)

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

+/ \(n+1=-1\Rightarrow n=-2.\)

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

+/ \(n+1=-2\Rightarrow n=-3.\)

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

Vậy \(n\in\left\{-2;0;-3;1\right\}\)

Answer 6

Ta có: n + 3 \(⋮\)n + 1

\(\Rightarrow\) n + 1 + 2 \(⋮\) n +1

\(\Rightarrow\) 2 \(⋮\) n+1

\(\Rightarrow\) n + 1 \(\in\) {1; -1; 2; -2}

\(\Rightarrow\) n \(\in\) {0; 1; -2; -3}

Vậy: n \(\in\) {0; 1; -2; -3}

Answer 7

Ta có: n+3 chia hết cho n+1

=> n+3 = n+1+2

=> n+1+2 chia hết cho n+1

=>2 chia hết cho n+1

Answer 8

Theo đề ta có : \(\left(n+3\right)⋮\left(n+1\right)\)

\(\Rightarrow\left[\left(n+1\right)+2\right]⋮n+1\)

\(\Rightarrow2⋮\left(n+1\right)\) ( do (n+1)\(⋮\)(n+1) )

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

\(\Rightarrow n\in\left\{-3;-2;0;1\right\}\)

Vậy \(n\in\left\{-3;-2;0;1\right\}\)