令 x/(x+1)=t
x=xt+t
x(1-t)=t
x=t/(1-t)
则f(x/(x+1))=f(t)=[t/(1-t)]2-2t/(1-t)-1
当t=0时
f(0)=[0/(1-0)]2-2x0/(1-0)-1
=-1
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