Gọi d = ƯCLN(2n + 1; 6n + 5)
=> \(\left\{{}\begin{matrix}2n+1⋮d\\6n+5⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left(2n+1\right)⋮d\\6n+5⋮d\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+5⋮d\end{matrix}\right.\)
<=> (6n + 5) - (6n + 3) \(⋮d\)
<=> 2 \(⋮d\)
Nếu d = 1 => (2n + 1; 6n + 5) = 1 => đpcm
Nếu d = 2 => 2n + 1 \(⋮2\) (loại)
@Trương Gia Kiệt
Gọi d \(\inƯC\left\{2n+1;6n+5\right\}\)
Ta có:
\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\6n+5⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3\left(2n+1\right)⋮d\\6n+5⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+5⋮d\end{matrix}\right.\)
=> (6n+5)-(6n+3) \(⋮\) d
\(\Rightarrow2⋮d\)
\(\Rightarrow d\inƯ\left(2\right)\)
Vì 2n+1 lẻ nên d chỉ có thể thuộc Ư(1)
=> d=1
=> (2n+1;6n+5)=1 (đpcm)
