Gọi d là \(ƯC\left(8n+5;6n+4\right)\)
⇒ \(\left\{{}\begin{matrix}8n+5⋮d\\6n+4⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left(8n+5\right)-\left(6n+4\right)⋮d\)
\(\Leftrightarrow\) \(\left(24n+15\right)-\left(24n+16\right)⋮d\) \(\Leftrightarrow-1⋮d\Leftrightarrow d=-1\)
⇒ \(ƯC\left(8n+5;6n+4\right)=-1\)
Vậy \(\dfrac{8n+5}{6n+4}\) là phân số tối giản
Đặt d=ƯCLN(8n+5,6n+4)
\(\Rightarrow\left\{{}\begin{matrix}8n+5⋮d\\6n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6\left(8n+5\right)⋮d\\8\left(6n+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\)2\(⋮\)d\(\Rightarrow\) d\(\in\)\(\left\{1,2\right\}\)
Mà 8n + 5 chẵn \(\Rightarrow\)d \(\ne\)2
\(\Rightarrow\)d=1 (đpcm)
\(\text{Gọi d=ƯCLN(8n+5;6n+4)}\left(d\in Z\right)\)
\(\Leftrightarrow8n+5⋮d\Leftrightarrow24n+15⋮d\)
\(\Leftrightarrow6n+4⋮d\Leftrightarrow24n+16⋮d\)
\(\Leftrightarrow1⋮d\)
\(\text{Vì }d\in Z;1⋮d\Leftrightarrow d=1\)
\(\Leftrightarrow\text{ƯCLN}\left(8n+5;6n+4\right)=1\)
\(\Rightarrow\text{ phân số }\dfrac{8n+5}{6n+4}\text{ tối giản với mọi }n\left(đpcm\right)\)