Question

已知a^2+4a+1=0,且(a^4-ma^2+1)/(2a^3+ma^2+2a)=3,求m的值

已知a^2+4a+1=0,且(a^4-ma^2+1)/(2a^3+ma^2+2a)=3,求m的值

Answer 1

由a^2+4a+1=0乘a^2得a^4+4a^2+1=0，乘a得a^3+4a^2+a=0
(a^4-ma^2+1)/(2a^3+ma^2+2a)=3
(a^4+4a^2+1-4a^2-ma^2)/(2a^3+8a^2+2a-8a^2+ma^2)=3
(-4a^2-ma^2)/(-8a^2+ma^2)=3
约去a^2得(-4-m)/(-8+m)=3
解得m=5

Answer 2

解：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

Answer 3

m=9.5