n+5 ⋮ n+1
=> n+1+4 ⋮ n+1
Vì n+1 ⋮ n+1 nên để n+1+4 ⋮ n+1 thì 4 ⋮ n+1
=> n+1 \(\in\) Ư(4) = {1;2;4}
n+1 | 1 | 2 | 4 |
n | 0 | 1 | 3 |
Vậy n = {0;1;3}
Ta có : \(n+5⋮n+1\) ; Mà : \(n+1⋮n+1\)
\(\Rightarrow\left(n+5\right)-\left(n+1\right)⋮n+1\Rightarrow n+5-n-1⋮n+1\)
\(\Rightarrow4⋮n+1\Rightarrow n+1\inƯ\left(4\right)\)
Mà : \(Ư\left(4\right)=\left\{1;2;4\right\}\) ; \(n+1\ge1\Rightarrow n+1\in\left\{1;2;4\right\}\Rightarrow n\in\left\{0;1;3\right\}\)
Vậy ...
n+5 chia cho n+1
ta co : n+5=n+1+5
vi n+1 chia het cho n+1 de n+1+4 chia het cho n+1
thi 4 chia het cho n+1 thuoc u(4)
ma u(4)={-4;-2;-1;1;2;4}
ta co n+1=-4
n =--5
n+1=-2
n =-1
n+1=-1
n =-2
n+1=1
n =0
n+1=2
n =1
n+1=4
n =3
vi de bai cho n la so tu nhien nen ta chon n=0 hoac n=1 hoac n=3
vay n=0 hoac n= 1 hoac n=3 thi 2chia het cho n+1
n+5\(⋮\)n+1
(n+1)+4\(⋮\)n+1
Vì n+1\(⋮\)n+1
Buộc 4\(⋮\)n+1=>n+1ϵƯ(4)={1;2;4}
Với n+1=1=>n=0
n+1=2=>n=1
n+1=4=>n=3
Vậy nϵ{0;1;3}