Question

若a^2+4a+1=0且(a^4+ma^2+1)/(3a^3+ma^2+3a)=5 求m的值

Answer 1

a^2+1=-4a so,a^4+2a^2+1=16a^2 so,a^4+1=14a^2

a^4+ma^2+1=(14+m)a^2

3a^3+ma^2+3a=3a(a^2+ma/3+1)=3a*(ma/3-4a)=(m-12)a^2

(a^4+ma^2+1)/(3a^3+ma^2+3a)=(14+m)/(m-12)=5

m=37/2