题目

设a=3x2﹣x+1，b=2x2+x，则（　　）                                                      A．a＞b                       B．a＜b                       C．a≥b                         D．a≤b   答案：C【考点】不等式比较大小．                                                                    【专题】计算题；不等式．                                                                      【分析】作差法化简a﹣b=x2﹣2x+1=（x﹣1）2≥0．                                        【解答】解：∵a=3x2﹣x+1，b=2x2+x，                                                   ∴a﹣b=x2﹣2x+1=（x﹣1）2≥0，                                                              ∴a≥b，                                                                                              故选：C．                                                                                          【点评】本题考查了作差法比较两个数的大小的应用．                                   　                                                                                                      

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