Question

(2n+5) chia hết cho (n+1)

Answer 1

( 2n + 5 ) ⋮ ( n + 1 ) <=> ( 2n + 2 ) + 3 ⋮ ( n + 1 ) <=> [ 2( n + 1 ) + 3 ] ⋮ ( n + 1 )

=> 3 ⋮ ( n + 1 ) => n + 1 thuộc ước của 3 => Ư(3) = { 1 ; 3 }

=> n + 1 = 1 => n = 0 ( tm )

=> n + 1 = 3 => n = 2 ( tm )

Vậy n = { 0 ; 2 }

Answer 2

Giải:
Ta có:

\(2n+5⋮n+1\)

\(\Rightarrow\left(2n+2\right)+3⋮n+1\)

\(\Rightarrow2\left(n+1\right)+3⋮n+1\)

\(\Rightarrow3⋮n+1\)

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

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

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

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

+) \(n+1=-3\Rightarrow n=-4\)

Vậy \(n\in\left\{0;-2;2;-4\right\}\)

Answer 3

cj giai theo pp ngắn gọn nhé

2n + 5 - 2n - 2 = 3

n+1(ư)3 = -1;-3;1;3

n = -2; -4; 0;2

Answer 4

(2n+5)\(⋮\)(n+1)

2n+2+3\(⋮\)n+1

2(n+1)+3\(⋮\)n+1

Vì 2(n+1)\(⋮\)n+1

Buộc 3\(⋮\)n+1=>n+1ϵƯ(3)={1;3}

Với n+1=1=>n=0

n+1=3=>n+2

Vậy nϵ{0;2}