Question

Chứng minh với mọi số tự nhiên n thì ƯCLN(21n+4;14n+3)=1.

Answer 1

Gọi \(d=ƯCLN\left(21n+4;4n+3\right)\) (\(d\in N\)*)

\(\Rightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Rightarrow1⋮d\)

Vì \(d\in N\)*;\(1⋮d\Rightarrow d=1\)

\(\RightarrowƯCLN\left(21n+4;14n+3\right)=1\rightarrowđpcm\)

Answer 2

Gọi d là ƯCLN (21n+4,14n+3)

Ta có: \(\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\left(21n+4\right)⋮d\\3\left(14n+3\right)⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\)

\(\Rightarrow1⋮d\)

Vì d \(\Rightarrow\) ƯCLN (21n+4,14n+3)

\(\Rightarrow\) d = 1

Vậy ƯCLN (21+4,14n+3) = 1 \(\forall\) n

Answer 3

Gọi d là UCLN(21n+4;14n+3)

\(\Leftrightarrow21n+4⋮d\Rightarrow2\left(21n+4\right)⋮d\Rightarrow42n+8⋮d\)

\(14n+3⋮d\Rightarrow3\left(14n+3\right)⋮d\Rightarrow42n+9⋮d\)

Vì:

\(42n+8;42n+9\)\(⋮d\)

Nên:

\(\left(42n+9\right)-\left(42n+8\right)⋮d\)

\(42n+9-42n-8⋮d\)

\(1⋮d\)

\(\Rightarrow\dfrac{21n+4}{14n+3}\)tối giản với mọi n