原函数在R上满足：f(x) = f(-x)向右平移一个单位后：另g(x) = f(x-1)而在R上g(x)是奇函数,所以：g(x)+g(-x) = 0而g(-x) = f(-x-1)所以：f(x-1) + f(-x-1) = 0有：f(x-1) + f(x+1) = 0 成立当x取2,6,10.2010时,f(1) +...f(x-1) + f(-x-1) = 0怎么得到f(x-1) + f(x+1) = 0因为f(x) = f(-x)g(x)=-g(-x)，那为什么g(-x) = f(-x-1)由g(x) = f(x-1)，g(-x)应该等于-f(x-1)啊晕死，怎么能这样理解？g(x) = f(x-1)则：g(t) = f(t-1)令t = -xg(-x) = f(-x-1)