Lời giải:
\(N=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a^2-2a+1)+15}=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a-1)^2+15}\)
Đặt \((a-1)^2=t\Rightarrow N=\frac{t^2-11t+30}{3t^2-18t+15}\)
\(=\frac{t^2-11t+30}{3(t^2-6t+5)}=\frac{(t-5)(t-6)}{3(t-1)(t-5)}\)
\(=\frac{t-6}{3(t-1)}=\frac{(a-1)^2-6}{3(a-1)^2-3}\)