Question

(难啊)函数题目：已知二次函数f(x)=a(a+1)x^2-(2a+1)x+1,a属于N*

Answer 1

1:用十字交叉法,

a -1

a+1 -1

f(x)=(ax-1)[(a+1)x-1]

x1=1/a,x2=1/(a+1)

所以,函数f(x)图像与x轴相交所截得的弦长为1/[a(a+1)]

2:

Ln=|xn1-xn2|=1/a-1/(a+1)

L1+L2+L3+L4+L5+L6=1/1-1/2+1/2-1/3+...+1/6-1/7=1-1/7=6/7