Gọi ƯCLN(2n+3,n+1)=d.
Ta có: \(2n+3⋮d\) ; \(n+1⋮d\) \(\Rightarrow2\left(n+1\right)⋮d\)
\(\Rightarrow2n+3-2\left(n+1\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1.\)
Vậy ƯCLN(2n+3,n+1)=1
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Gọi d là UCLN(2n+3;n+1)
Theo đề bài ta có:
\(2n+3⋮d\)
\(n+1⋮d\Rightarrow2\left(n+1\right)⋮d\Rightarrow2n+2⋮d\)
\(\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d\)
\(2n+3-2n-2⋮d\)
\(1⋮d\)
\(d_{MAX}\Rightarrow d=1\)
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