设f(x)=(x-1)/(x+1),fn(x)=f{f[f··f(x)]}一共n个f,则f2006(x)=

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设f(x)=(x-1)/(x+1),fn(x)=f{f[f··f(x)]}一共n个f,则f2006(x)=

设f(x)=(x-1)/(x+1),fn(x)=f{f[f··f(x)]}一共n个f,则f2006(x)=
设f(x)=(x-1)/(x+1),fn(x)=f{f[f··f(x)]}一共n个f,则f2006(x)=

设f(x)=(x-1)/(x+1),fn(x)=f{f[f··f(x)]}一共n个f,则f2006(x)=
fn(x)是个周期函数f2(x)=-1/x,f3(x)=-(x+1)/(x-1),f4(x)=x
所以T=4
f2006(x)=f2(x)=-1/x

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