易知x^2=x+1a^2=a+1a^4=a^2+2a+1 =a+1+2a+1 =3a+2a^4+3b=3(a+b)+2韦达定理得原式=3*(-1)+2=-1
由题知a^2-a=1,a+b=1.则a^4+3b=a^4-a^2+a^2+3b=(a^2+a)(a^2-a)+a^2-a+a+3b=a^2+a+1+a+3b=a^2-a+3(a+b)=1+3=4
同意Namedgod 的
