b) \(\frac{n^3+2n}{n^4+3n^2+1}\)
Giải:
Gọi \(ƯCLN\left(n^3+2n;n^4+3n^2+1\right)\) là \(d\)
\(\Rightarrow\left\{\begin{matrix}n^3+2n⋮d\\n^4+3n^2+1⋮d\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n\left(n^3+2n\right)⋮d\\n^4+2n^2⋮d\end{matrix}\right.\)
Do đó:
\(\left(n^4+3n^2+1\right)-\left(n^4+2n^2\right)⋮d\) Hay \(n^2+1⋮d\) (1)
\(\Rightarrow\left(n^2+1\right)\left(n^2+1\right)⋮d\) Hay \(n^4+2n^2+1⋮d\)
\(\Rightarrow\left(n^4+3n^2+1\right)-\left(n^4+2n^2+1\right)⋮d\) Hay \(n^2⋮d\) (2)
Từ (1) và (2)
\(\Rightarrow\left(n^2+1\right)-n^2⋮d\) Hay \(1⋮d\)
\(\RightarrowƯCLN\left(n^3+2n;n^4+3n^2+1\right)=1\) hoặc \(-1\)
\(\Rightarrow\frac{n^3+2n}{n^4+3n^2+1}\) là phân số tối giản (Đpcm)
