n+3\(⋮\)n+1
=> (n+1)+2\(⋮\)n+1
=>2\(⋮\)n+1
=> n+1\(\in\)Ư(2)={+-1;+-2}
Do n la so tu nhien nen n+1 cung la so tu nhien
Ta co bang sau
n+1 | 1 | 2 |
n | 0 | 1 |
Vay la n=0;1
Ta có : n + 3 \(⋮\)n+1
=> (n+1) + 2 \(⋮\)n+1
Mà n+1 \(⋮\)n+1
=> 2\(⋮\)n+1
Do \(n\in N\)
=> n+1 \(\in\)Ư(2) = {1 ; 2}
=> n = 0;1
Vậy n = 0;1