Question

已知：a²+4a+1=0，且

a4+ma2+1

2a3+ma2+2a

=3，则m的值为

 

．

Answer 1

分析：由a²+4a+1=0，得a²=-4a-1，代入所求的式子化简即可．

解答：解：∵a²+4a+1=0，∴a²=-4a-1，

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

=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．

点评：解题关键是两次用到了整体代入的思想，它在解题中起到了降幂，从而化难为易的作用．