Question

已知a²+3a+1=0,求3a³+(a²+5)(a²-1)-a(5a+6)的值

Answer 1

a²+3a+1=0
a²=-3a-1
以下a²都用-3a-1代替

所以a³=a(a²)
=a(-3a-1)
=-3a²-a
=-3(-3a-1)-a
=8a+3

原式=3(8a+3)+(-3a-1+5)(-3a-1-1)-5a²-6a
=24a+9+(3a-4)(3a+2)-5(-3a-1)-6a
=24a+9+9a²-6a-8+15a+5-6a
=27a+6+9(-3a-1)
=27a+6-27a-9
=-3

Answer 2

由顶点
y=a(x+2)²+1
开口向下则a<0
假设a=-1
则y=-x²-4x-3
1、
1/[n(n+1)]=1/n-1/(n+1)

2、
1/2+1/6+1/12+…+1/2005*2006+1/2006*2007
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/2006-1/2007
=1-1/2007
=2006/2007