题目内容
已知：a2+4a+1=0，且
=3，则m的值为______．
| a4+ma2+1 |
| 2a3+ma2+2a |
∵a2+4a+1=0，∴a2=-4a-1，
=
=
=
=
=3
即（-56-4m）a-14-m=（-12m+96）a-3m+24，
∴-56-4m=-12m+96，-14-m=-3m+24，
解得m=19．
故答案为19．
| a4+ma2+1 |
| 2a3+ma2+2a |
| (-4a-1)2+ma2+1 |
| 2a(-4a-1)+ma2+2a |
=
| (16+m)a2+8a+2 |
| (m-8)a2 |
=
| (16+m)(-4a-1)+8a+2 |
| (m-8)(-4a-1) |
=
| (-56-4m)a-14-m |
| (-4m+32)a-m+8 |
即（-56-4m）a-14-m=（-12m+96）a-3m+24，
∴-56-4m=-12m+96，-14-m=-3m+24，
解得m=19．
故答案为19．
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