题目

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） 答案：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．                 

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