已知a^2+4a+1=0,且(a^4-ma^2+1)/(2a^3+ma^2+2a)=3,求m的值
问题描述：
已知a^2+4a+1=0,且(a^4-ma^2+1)/(2a^3+ma^2+2a)=3,求m的值
答
a^2=-4a-1 a^3=a^2*a=-4a^2-a=-4(-4a-1)-a=16a+4-a=15a+4 a^4=(a^2)^2=16a^2+8a+1=16(-4a-1)+8a+1=-56a-15 (a^4+ma^2+1)/2a^3+ma^2+2a=3 a^4+ma^2+1=6a^3+3ma^2+6a 2ma^2=-6a^3-6a+a^4+1 -8am-2m=-90a-24-6a-56a-15+...