Question

若實數a，b滿足a+b²=1，則a²+b²的最小值是　　　　　　．                       

Answer 1

　．                                                                                                

【考點】配方法的應用；非負數的性質：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+≥，                                                        

∴當a=時，a²+b²有最小值．                                                              

故答案為．                                                                                      

【點評】本題考查了配方法的應用，非負數的性質，將b²=1﹣a代入得到a²+b²=a²+1﹣a是解題的關鍵．