\(\frac{n^7+n^2+1}{n^8+n+1}=\frac{\left(n^2+n+1\right)\left(n^5-n^4+n^2-n+1\right)}{\left(n^2+n+1\right)\left(n^6-n^5+n^3-n^2+1\right)}=\frac{n^5-n^4+n^2-n+1}{n^6-n^5+n^3-n^2+1}\)

=>phân số ban đầu chưa tối giản với mọi n

Ta có :

\(\frac{n^7+n^2+1}{n^8+n+1}=\frac{n^7-n^4+n^4-n+n^2+n+1}{n^8-n^5+n^5-n^2+n^2+n+1}\)

\(=\frac{n^4\left(n^3-1\right)+n\left(n^3-1\right)+\left(n^2+n+1\right)}{n^5\left(n^3-1\right)+n^2\left(n^3-1\right)+\left(n^2+n+1\right)}\)

\(=\frac{n^4\left(n-1\right)\left(n^2+n+1\right)+n\left(n-1\right)\left(n^2+n+1\right)+\left(n^2+n+1\right)}{n^5\left(n-1\right)\left(n^2+n+1\right)+n^2\left(n-1\right)\left(n^2+n+1\right)+\left(n^2+n+1\right)}\)

\(=\frac{\left(n^2+n+1\right)\left(n^5-n^4+n^2-n+1\right)}{\left(n^2+n+1\right)\left(n^6-n^5+n^3-n+1\right)}\)

\(=\frac{n^5-n^4+n^2-n+1}{n^6-n^5+n^3-n+1}\)

Do phân số \(\frac{n^7+n^2+1}{n^8+n+1}\) còn thu gọi được thành \(\frac{n^5-n^4+n^2-n+1}{n^6-n^5+n^3-n+1}\) nên nó chưa tối giản (đpcm)

mỗi câu mà đăng hoài zậy @@

Em chưa học làm dạng này , em làm thử thôi nhá, sai xin chỉ dạy thêm nha

2 . \(\dfrac{n^7+n^2+1}{n^8+n+1}=\dfrac{n^7-n+n^2+n+1}{n^8-n^2+n^2+n+1}\)

\(=\dfrac{n\left(n^6-1\right)+n^2+n+1}{n^2\left(n^6-1\right)+n^2+n+1}=\dfrac{n\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}\)\(=\dfrac{n\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}\)

\(=\dfrac{\left(n^2+n+1\right)\left[\left(n^4+n\right)\left(n-1\right)\right]}{\left(n^2+n+1\right)\left[\left(n^5+n^2\right)\left(n-1\right)+1\right]}\)

\(=\dfrac{n^5-n^4+n^2-n}{n^6-n^5+n^3-n^2+1}=\dfrac{n^4\left(n-1\right)+n\left(n-1\right)}{n^5\left(n-1\right)+n^2\left(n-1\right)+1}\)

\(=\dfrac{\left(n-1\right)\left(n^4+n\right)}{\left(n-1\right)\left(n^5+n^2\right)+1}\)

Vậy ,với mọi số nguyên dương n thì phân thức trên sẽ không tối giản