Question

Chứng minh rằng:

\(\frac{n^2+n+1}{n^4+n^2+1}\) ( n thuộc N*) không là phân số tối giản

Answer 1

\(\frac{n^2+n+1}{n^4+n^2+1}=\frac{n^2+n+1}{\left(n^2+n+1\right)+\left(n^4-n\right)}=\frac{n^2+n+1}{\left(n^2+n+1\right)+n\left(n^3-1\right)}=\frac{n^2+n+1}{\left(n^2+n+1\right)+n\left(n-1\right)\left(n^2+n+1\right)}=\frac{n^2+n+1}{\left(n^2+n+1\right)\left(n^2-n+1\right)}=\frac{1}{n^2-n+1}\)

Answer 2

Nguyễn Trần Nhã Anh cách biến đổi khác dễ hơn :)
\(\frac{n^2+n+1}{n^4+n^2+1}=\frac{n^2+n+1}{n^4+2n^2+1-n^2}=\frac{n^2+n+1}{\left(n^2+1\right)-n^2}=\frac{n^2+n+1}{\left(n^2-n+1\right)\left(n^2+n+1\right)}=\frac{1}{n^2-n+1}\)