Đặt \(ƯCLN\left(12n+1;30n+2\right)=d\)
\(\Rightarrow\left\{\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\Rightarrow\left\{\begin{matrix}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}60n+5⋮d\\60n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy phân số \(A=\frac{12n+1}{30n+2}\) là phân số tối giản (Đpcm)
Để \(\frac{12n+1}{30n+2}\) tối giản thì ƯCLN(12n+1; 30n+2) = 1
Đặt d = ƯCLN(12n+1; 30n+2)
\(\Leftrightarrow\left\{\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{matrix}\right.\Leftrightarrow}\left\{\begin{matrix}60n+5⋮d\\60n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) (60n + 5) - (60n + 4) \(⋮\) d
\(\Rightarrow\) 60n + 5 - 60n - 4 \(⋮\) d
\(\Rightarrow\) 1 \(⋮\) d \(\Rightarrow\) d = 1
Vậy \(\frac{12n+1}{30n+2}\) tối giản (chứng minh xong).
