Question

已知函数f(x)=x/(x+1)+(x+1)/(x+2)+(x+2)/(x+3)+(x+3)/(x+4)

则f(-5/2+根号2)+f(-5/2-根号2)=

Answer 1

f(x)=4-1/(x+1)-1/(x+2)-1/(x+3)-1/(x+4) 1/(x+x1+1)+1/(x-x1+1)=2*(x+1)/[(x+1)^2-x1^2]
f（x+x1）+f(x-x1)=8-2*(x+1)/[(x+1)^2-x1^2]-2*(x+2)/[(x+2)^2-x1^2]-2*(x+3)/[(x+3)^2-x1^2]-2*(x+4)/[(x+4)^2-x1^2]
将x=-5/2，x1=根号2代入
f(-5/2+根号2)+f(-5/2-根号2)=8+12-4/7+4/7-12=8

Answer 2

先观察到-5/2+根号2+-5/2-根号2=常数-5，猜想x和-5-x的函数值和为常数，然后验证可得。