Gọi d là ƯCLN ( 2n + 5, 3n + 7 )
\(2n+5⋮d\Rightarrow3.\left(2n+5\right)⋮d\Rightarrow6n+15⋮d\)
\(3n+7⋮d\Rightarrow2.\left(3n+7\right)⋮d\Rightarrow6n+14⋮d\)
\(\left[\left(6n+15\right)-\left(6n+14\right)\right]⋮d\)
\(\left[6n+15-6n-14\right]⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
PP/ss: Hoq chắc ạ ((:
Gọi ƯCLN(2n+5,3n+7)=d
Ta có:2n+5:d,suy ra 3(2n+5)=6n+15:d
3n+7:d suy ra 2(3n+7)=6n+14:d
Vậy (6n+15)-(6n+14):d
Nên 1:d,suy ra 2 ƯCLN của 2 số là 1
--hok tốt--
Gọi d = UCLN ( 2n+5 ; 3n + 7 )
\(\Rightarrow2n+5⋮d;3n+7⋮d\)
\(\Rightarrow3\left(2n+5\right)⋮d\)
\(2\left(3n+7\right)⋮d\)
\(\Rightarrow6n+15⋮d\) \(6n+14⋮d\)
\(\Rightarrow\left(6n+15\right)-\left(6n+14\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\)
mà \(n\in N\Rightarrow d=1\)
\(\Rightarrow UCLN\left(2n+5;3n+7\right)=1\)