15 ⋮ (2x + 1)
Nhận xét: vì x là số tự nhiên => x >0 => 2x + 1 >0
Ta có:
2x + 1 ϵ Ư(15) = {1; 3; 5; 15}
=>ta có bảng:
2x + 1 | 1 | 3 | 5 | 15 |
x | 0 | 1 | 2 | 7 |
=>x ϵ {0; 1; 2; 7}
\(15⋮\left(2x+1\right)\)
=> \(2x+1\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
=> \(\left[{}\begin{matrix}2x+1=1\\2x+1=3\\2x+1=5\\2x+1=15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x=1-1=0\\2x=3-1=2\\2x=5-1=4\\2x=15-1=14\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0:2=0\\x=2:2=1\\x=4:2=2\\x=14:2=7\end{matrix}\right.\)
Vậy x ∈ { 0 ; 1 ; 2 ; 7 }