Gọi ƯC(n+3;2n+5) là d
Ta có:
\(\left\{{}\begin{matrix}\left(n+3\right)⋮d\\\left(2n+5\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2.\left(n+3\right)⋮d\\\left(2n+5\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2n+6⋮d\\2n+5⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\)
\(\Rightarrow\) \(\left(2n+6-2n-5\right)⋮d\)
\(\Rightarrow\) 1 \(⋮\)d
\(\Rightarrow\) d = 1
Vậy ước chung của 2 số n + 3 và 2n + 5 là 1
Gọi \(UC_{\left(n+3;2n+5\right)}=d\left(d\in N\right).\)
\(\Rightarrow\left\{{}\begin{matrix}n+3⋮d.\\2n+5⋮d.\end{matrix}\right.\)
\(\Rightarrow\left(n+3\right)-\left(2n+5\right)⋮d.\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d.\)
\(\Rightarrow1⋮d.\)
mà \(d\in N.\)
\(\Rightarrow d=1.\)
Vậy \(UC_{\left(n+3;2n+5\right)}=1.\)
Gọi UC(n+3;2n+5) là d
\(\Rightarrow\left\{{}\begin{matrix}n+3⋮d\\2n+5⋮d\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2n+6⋮d\\2n+5⋮d\end{matrix}\right.\\ \Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\\ \Rightarrow1⋮d\\ \Rightarrow d=1\)
VậyUC(n+3;2n+5) là 1