Question

已知f(根号x+1)=x+2根号x,求f(x+1)的解析式

Answer 1

设√x+1=t

x=(t-1)²

f(√x+1)=f(t)=(t-1)²+2(t-1)

=t²-2t+1+2t-2

=t²-1

所以f(x)=x²-1

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