Question

[Hoàng Luke]

Chứng minh phân số sau đây là phân số tối giản :\(\dfrac{n^3+2n}{n^4+3n^2+1}\)

Answer 1

[Nguyễn Thanh Hằng]

Gọi \(d=ƯCLN\left(n^3+2n;n^4+3n^2+1\right)\left(d\in N\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^3+2n⋮d\\n^4+3n^2+1⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^4+2n^2⋮d\\n^4+3n^2+1⋮d\end{matrix}\right.\)

\(\Leftrightarrow n^2+1⋮d\)

Mà \(n^3+2n⋮d\)

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

\(\Leftrightarrow n⋮d\)

Mà \(n^2+1⋮d\)

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

\(\Leftrightarrow1⋮d\)

Vì \(d\in N;1⋮d\Leftrightarrow d=1\)

\(\LeftrightarrowƯCLN\left(n^3+2n;n^4+3n^2+1\right)=1\)

\(\Leftrightarrow\) Phân số \(\dfrac{n^3+2n}{n^4+3n^2+1}\) tối giản với mọi n

Answer 2

[Chippy Linh]

Gọi \(d=ƯCLN\left(n^3+2n;n^4+3n^2+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n^3+2n⋮d\\n^4+3n^2+1⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n^4+2n^2⋮d\\n^4+3n^2+1⋮d\end{matrix}\right.\)

\(\Rightarrow n^2+1⋮d\)

Mà n³ + 2n \(⋮\) d

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

\(\Rightarrow n⋮d\)

Mà n² + 1 \(⋮\) d

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

\(\Rightarrow1⋮d\)

Vì \(d\in N;1⋮d\Rightarrow d=1\)

\(\RightarrowƯCLN\left(n^3+2n;n^4+3n^2+1\right)=1\)

Vậy phân số \(\dfrac{n^3+2n}{n^4+3n^2+1}\) tối giản \(\forall n\in N\) => đpcm