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