原函數(shù)在R上滿足：f(x) = f(-x)向右平移一個單位后：另g(x) = f(x-1)而在R上g(x)是奇函數(shù),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 成立當x取2,6,10.2010時,f(1) +...f(x-1) + f(-x-1) = 0怎么得到f(x-1) + f(x+1) = 0因為f(x) = f(-x)g(x)=-g(-x)，那為什么g(-x) = f(-x-1)由g(x) = f(x-1)，g(-x)應該等于-f(x-1)啊暈死，怎么能這樣理解？g(x) = f(x-1)則：g(t) = f(t-1)令t = -xg(-x) = f(-x-1)