(3x+13)\(⋮\) (3x+1)
Ta có : 3x+13 = 3x+1+12
mà (3x+1) \(⋮\)(3x+1) để (3x+13)\(⋮\) (3x+1)
thì => 12 \(⋮\) (3x+1) hay 3x+1 \(\in\) Ư(12)={1;2;3;4;6;12}
ta có bảng sau
3x+1 | 1 | 2 | 3 | 4 | 6 | 12 |
x | 0 | / | / | 1 | / | / |
vậy x\(\in\){0;1}
Cách đơn giản hơn :))
\(3x+13⋮3x+1\)
\(3x+1+12⋮3x+1\)
\(12⋮3x+1\)hay \(3x+1\inƯ\left(12\right)=\left\{1;2;3;4;12\right\}\)
3x + 1 | 1 | 2 | 3 | 4 | 12 |
3x | 0 | 1 | 2 | 3 | 11 |
x | 0 | 1/3 | 2/3 | 1 | 11/3 |
Vì \(n\in N\)Suy ra n = 0 ; 3