Question

已知a^2+4a+1=0,并且有（a^4+ma^2+1)/(2a^3+ma^2+2a)=3,求m的值？

求求好。不能答出来我啊

Answer 1

a^2=-4a-1
a^4=(-4a-1)^2=16a^2+8a+1=16(-4a-1)+8a+1=-56a-15
a^3=a^2*a=a(-4a-1)=-4a^2-a=-4(-4a-1)-a=15a+4

(a^4+ma^2+1)/(2a^3+ma^2+2a)=3
a^4+ma^2+1=6a^3+3ma^2+6a
所以-56a-15+m(-4a-1)+1=6(15a+4)+3m(-4a-1)+6a
m(-4a-1+12a+3)=90a+6a+56a+24+15-1
m(8a+2)=152a+38
2m(4a+1)=38(4a+1)
m=19