f(x)=(1+x)/(1-x),且f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=
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![f(x)=(1+x)/(1-x),且f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=](/uploads/image/z/10905528-48-8.jpg?t=f%28x%29%3D%281%2Bx%29%2F%281-x%29%2C%E4%B8%94f1%28x%29%3Df%28x%29%2Cfk%2B1%28x%29%3Df%28fk%28x%29%29%2Ck%3D1%2C2%2C%E2%80%A6%2C%E5%88%99f2009%28x%29%3D)
f(x)=(1+x)/(1-x),且f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=
f(x)=(1+x)/(1-x),且f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=
f(x)=(1+x)/(1-x),且f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=
f1(x)=(1+x)/(1-x),f2(x)=-1/x,f3(x)=(x-1)/1+x,
f4(x)=x,f5(x)=(1+x)/(1-x),所以周期为4
所以f2009(x)=f1(x)=(1+x)/(1-x)