题目

若实数a，b满足a+b2=1，则a2+b2的最小值是　　　　　　．                        答案：　．                                                                                                 【考点】配方法的应用；非负数的性质：偶次方．                                          【分析】由a+b2=1，得出b2=1﹣a，代入得到a2+b2=a2+1﹣a，利用配方法即可求解．                  【解答】解：∵a+b2=1，                                                                          ∴b2=1﹣a，                                                                                        ∴a2+b2=a2+1﹣a=（a﹣）2+≥，                                                         ∴当a=时，a2+b2有最小值．                                                               故答案为．                                                                                       【点评】本题考查了配方法的应用，非负数的性质，将b2=1﹣a代入得到a2+b2=a2+1﹣a是解题的关键．                 　                                                                                                      

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