Do a là nghiệm nên \(a^2-3a+1=0\Rightarrow a^2=3a-1\)
\(\Rightarrow a^4=\left(3a-1\right)^2=9a^2-6a+1=9\left(3a-1\right)-6a+1=21a-8\)
\(Q=\frac{a^2}{a^4+a^2+1}=\frac{3a-1}{21a-8+3a-1+1}=\frac{3a-1}{24a-8}=\frac{3a-1}{8\left(3a-1\right)}=\frac{1}{8}\)