a, Gọi \(d=ƯCLN\left(n,n+1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}n⋮d\\n+1⋮d\end{matrix}\right.\)
\(\Leftrightarrow1⋮d\)
\(\Leftrightarrow d=1\)
\(\LeftrightarrowƯCLN\left(n,n+1\right)=1\)
b, Ta có :
\(ƯCLN\left(n,n+1\right)=1\left(cmt\right)\)
\(\Leftrightarrow n+1;n\) nguyên tố cùng nhau
\(\Leftrightarrow BCNN\left(n+1;n\right)=\left(n+1\right)n=n^2+n\)
a, Gọi d=ƯCLN(n,n+1)d=ƯCLN(n,n+1)
=> \(\left\{{}\begin{matrix}n⋮d\\n+1⋮d\end{matrix}\right.\)
⇔{n⋮dn+1⋮d
⇔1⋮d⇔1⋮d
⇔d=1⇔d=1
⇔ƯCLN(n,n+1)=1⇔ƯCLN(n,n+1)=1
b, Ta có :
ƯCLN(n,n+1)=1(cmt)ƯCLN(n,n+1)=1(cmt)
⇔n+1;n⇔n+1;n nguyên tố cùng nhau
⇔BCNN(n+1;n)=(n+1)n=n^2+n
