已知a²+4a+1=0,且（a四次方+ma²+1）÷（2a³+ma²+2a）=3,求m的值
问题描述：
已知a²+4a+1=0,且（a四次方+ma²+1）÷（2a³+ma²+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+1=-152a-38
2m(4a+1)=38(4a+1)
m=19