已知f（）=，则（ ）                               

已知f（）=，则（　　）                                                         

A．f（x）=x²+1（x≠0）                                  B．f（x）=x²+1（x≠1）    C．f（x）=x²﹣1（x≠1）    D．f（x）=x²﹣1（x≠0）

C【考点】函数解析式的求解及常用方法．                                              

【专题】转化思想；换元法；函数的性质及应用．                                         

【分析】由f（）=，变形为=﹣1，即可得出．                  

【解答】解：由，            

得f（x）=x²﹣1，                                                                              

又∵≠1，                                                                                     

∴f（x）=x²﹣1的x≠1．                                                                         

故选：C．                                                                                         

【点评】本题考查了函数的解析式求法，考查了推理能力与计算能力，属于基础题．