-Ta có: \(n^4+n^2+1=\left(n^4+n^3+n^2\right)+\left(-n^3-n^2-n\right)+\left(n^2+n+1\right)=n^2\left(n^2+n+1\right)-n\left(n^2+n+1\right)+\left(n^2+n+1\right)=\left(n^2+n+1\right)\left(n^2-n+1\right)\)
\(\Rightarrow\dfrac{n^2+n+1}{n^4+n^2+1}=\dfrac{n^2+n+1}{\left(n^2+n+1\right)\left(n^2-n+1\right)}=\dfrac{1}{n^2-n+1}\).
-Vậy \(\dfrac{n^2+n+1}{n^4+n^2+1}\left(n\in Nsao\right)\) không là phân số tối giản.