已知a2+4a+1=0且(a2+ma2+1)/(3a3+ma2+3a)=5,则m=?好的我会提高悬赏分!对不起,弄错了,是已知a2+4a+1=0且(a4+ma2+1)/(3a3+ma2+3a)=5,则m=?

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已知a2+4a+1=0且(a2+ma2+1)/(3a3+ma2+3a)=5,则m=?好的我会提高悬赏分!对不起,弄错了,是已知a2+4a+1=0且(a4+ma2+1)/(3a3+ma2+3a)=5,则m=?

已知a2+4a+1=0且(a2+ma2+1)/(3a3+ma2+3a)=5,则m=?好的我会提高悬赏分!对不起,弄错了,是已知a2+4a+1=0且(a4+ma2+1)/(3a3+ma2+3a)=5,则m=?
已知a2+4a+1=0且(a2+ma2+1)/(3a3+ma2+3a)=5,则m=?好的我会提高悬赏分!
对不起,弄错了,是已知a2+4a+1=0且(a4+ma2+1)/(3a3+ma2+3a)=5,则m=?

已知a2+4a+1=0且(a2+ma2+1)/(3a3+ma2+3a)=5,则m=?好的我会提高悬赏分!对不起,弄错了,是已知a2+4a+1=0且(a4+ma2+1)/(3a3+ma2+3a)=5,则m=?
a²+1=-4a
平方
a^4+2a²+1=16a²
a^4+1=14a²
则(14a²+ma²)/{a[3(a²+1)+ma]}=5
a²(m-4)/[a(-12a+ma)]=5
a²(m-4)/[a²(-12+m)]=5
(m-4)/(-12+m)=5
m-4=-60+5m
m=14

a^2+1=-4a so, a^4+2a^2+1=16a^2 so, a^4+1=14a^2
a^4+ma^2+1=(14+m)a^2
3a^3+ma^2+3a=3a(a^2+ma/3+1)=3a*(ma/3-4a)=(m-12)a^2
(a^4+ma^2+1)/(3a^3+ma^2+3a)=(14+m)/(m-12)=5
m=37/2

∵a2+4a+1=0,∴a2=-4a-1,
a4+ma2+13a3+ma2+3a=(-4a-1)2+ma2+13a(-4a-1)+ma2+3a
=(16+m)a2+8a+2(m-12)a2
=(16+m)a2+8a+2(m-12)(-4a-1)
=(16+m)(-4a-1)+8a+2(m-12)(-4a-1)=5,
∴(16+m)(-4a-1)+8a+2=5...

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∵a2+4a+1=0,∴a2=-4a-1,
a4+ma2+13a3+ma2+3a=(-4a-1)2+ma2+13a(-4a-1)+ma2+3a
=(16+m)a2+8a+2(m-12)a2
=(16+m)a2+8a+2(m-12)(-4a-1)
=(16+m)(-4a-1)+8a+2(m-12)(-4a-1)=5,
∴(16+m)(-4a-1)+8a+2=5(m-12)(-4a-1),
原式可化为(16+m)(-4a-1)-5(m-12)(-4a-1)=-8a-2,
即[(16+m)-5(m-12)](-4a-1)=-8a-2,
∵a≠0,
∴(16+m)-5(m-12)=2,
解得m=372.
故答案为372.

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