Question

Cho n là số tự nhiên, ƯCLN(2n+3; 3n+4) là ...

Answer 1

Gọi ƯCLN(2n+3,3n+4) là d

Ta có: \(2n+3⋮d\Rightarrow3\left(2n+3\right)⋮d\Rightarrow6n+9⋮d\)

\(3n+4⋮d\Rightarrow2\left(3n+4\right)⋮d\Rightarrow6n+8⋮d\)

\(\Rightarrow6n+9-\left(6n+8\right)⋮d\)

\(\Rightarrow6n+9-6n-8⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow UCLN\left(2n+3,3n+4\right)=1\)

Answer 2

Gọi d = ƯCLN (2n+3; 3n+4)

Ta có :

\(\frac{2n+3⋮d}{3n+4⋮d}=>\frac{3\left(2n+3\right)}{2\left(3n+4\right)}=>\frac{6n+9}{6n+8}\)

=> \(\left(6n+9\right)-\left(6n+8\right)⋮d\)

=> \(1⋮d\) =>\(d=1\)

Vậy ƯCLN (2n+3;3n+4) =1