Question

设函数y=f(x)是定义域为R的函数,且f(x+2)[1-f(x)]=1+f(x),又f(2)=2+根号2,则f(200

Answer 1

f(x+2)=[1+f(x)]/[1-f(x)]

f(4)=(3+根号2)/(-1-根号2)=1-2根号2

f(6)=(根号2-2)/2

f(8)=(2根号2+1)/7

f(10)=2+根号2

所以周期为8,即f(x+8)=f(x)

f(2008)=f(250*8+8)=f(8)=(2根号2+1)/7