Question

Chứng tỏ rằng với mọi số tự nhiên n,các số là hai nguyên tố cùng nhau:

a,n+1 và 2n+3 ;b, 3n+2 và 5n+3

Answer 1

a) Gọi \(d\) là \(UCLN\left(n+1;2n+3\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n+1⋮d\\2n+3⋮d\end{matrix}\right.\)

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

\(\Rightarrow2n+3-2n-2⋮d\)

\(\Rightarrow1⋮d\Leftrightarrow d=1\)

Vậy \(\left(n+1;2n+3\right)=1\rightarrowđpcm\)

b) Gọi \(d\) là \(UCLN\left(3n+2;5n+3\right)\)

\(\Rightarrow\left\{{}\begin{matrix}3n+2⋮d\\5n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}15n+10⋮d\\15n+9⋮d\end{matrix}\right.\)

\(\Rightarrow15n+10-15n-9⋮d\)

\(\Rightarrow1⋮d\Leftrightarrow d=1\)

Vậy \(\left(3n+2;5n+3\right)=1\rightarrowđpcm\)

Answer 2

Goi d la UCLN(n+1;2n+3)

=>{n=1:'d

{2n+3:'d

=>{2n+2:'d

{2n+3:'d

=>2n+3-2n-2:'d

1:d(=)d=1

Vay (n+1;2n+3)=1=>dpcm

b,Goi d la UCLN(3n+2;5n+3)

=>{3n+2:d

{5n+3:d

=>{15n+10:d

{15n+9:d

=>15n+10-15n-9:d

=1:d(=)d=1

Vay(3n+2;5n+3)=1=>dpcm