已知f1(x)=(2x-1)/(x+1),fn+1(x)=f1[fn(x)](n=1,2,3,……),求f30（x）

f2(x)={2[(2x-1)/(x+1)]-1}/{[(2x-1)/(x+1)]+1}
=(x-1)/x
f3(x)={2[(x-1)/x]-1}/{[(x-1)/x]+1}
=(x-2)/(2x-1)
f4(x)={2[(x-2)/(2x-1)]-1}/{[(x-2)/(2x-1)]+1}
=-1/(x-1)
f5(x)={2[-1/(x-1)]-1}/{[-1/(x-1)]+1}
=(-x-1)/(x-2)
f6(x)={2[(-x-1)/(x-2)]-1}/{[(-x-1)/(x-2)]+1}
=x
f7(x)=(2x-1)/(x+1)=f1(x)
所以从f1(x)到f6(x)每6个一循环
30=4*6+6
所以f30(x)=f6(x)=x