Question

Chứng tỏ rằng : ƯCLN(n+2,2n+3)=1

 

Answer 1

Gọi \(d=\left(n+2;2n+3\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n+2⋮d\\2n+3⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+4⋮d\\2n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow\)\(1⋮d\Rightarrow d=1\)

Answer 2

Gọi d là \(UCLN\left(n+2,2n+3\right)\), khi đó:

\(\left\{{}\begin{matrix}n+2⋮d\\2n+3⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2n+4⋮d\\2n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\)

\(\Rightarrow1⋮d\)

Vậy \(UCLN\left(n+2,2n+3\right)=1\) (dpcm)