Gọi ƯCLN(12n + 1,30n + 2) = d
Ta có: \(12n+1⋮d\Rightarrow5\left(12n+1\right)⋮d\Rightarrow60n+5⋮d\)
\(30n+2⋮d\Rightarrow2\left(30n+2\right)⋮d\Rightarrow60n+4⋮d\)
=> 60n + 5 - (60n + 4) \(⋮\) d
=> 60n + 5 - 60n - 4 \(⋮\) d
=> 1 \(⋮\) d => d = 1
=> \(\frac{12n+1}{30n+2}\) là phân số tối giản
Đặt (12n+1, 30n+2) = d
\(\Rightarrow\) \(\left\{\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\) \(\Rightarrow\)\(\left\{\begin{matrix}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left\{\begin{matrix}60n+5⋮d\\60n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) ( 60n + 5) - (60n +4) \(⋮\) d
\(\Rightarrow\) 5 - 1 \(⋮\) d
\(\Rightarrow\) 1 \(⋮\) d
\(\Rightarrow\) d = 1
Vậy \(\frac{12n+1}{30n+2}\) là phân số tối giản.
