f(x/(x+1))=x2-2x-1,求f(0)
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![f(x/(x+1))=x2-2x-1,求f(0)](/uploads/image/z/2535009-33-9.jpg?t=f%28x%2F%28x%2B1%29%29%3Dx2-2x-1%2C%E6%B1%82f%280%29)
f(x/(x+1))=x2-2x-1,求f(0)
f(x/(x+1))=x2-2x-1,求f(0)
f(x/(x+1))=x2-2x-1,求f(0)
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
答案应该是-1
x/(x+1)=0 则x=0
f(0)=0+0-1=-1