已知a2+4a+1=0,且a4+ma2+13a3+ma2+3a=5,则m=_.

问题描述:

已知a2+4a+1=0,且

a4+ma2+1
3a3+ma2+3a
=5,则m=______.

∵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(...