1.问题的提出 例1 如果下列三个方程x~2+4ar-4a+3=0,x~2+(a-1)x+a~2=0,x~2+2ax-2a=0中至少有一个方程有实根,求实数a的取值范围。 分析:正面理解题目中的关键词“至少”,可得如下三类: (1)只有一个方程有实根,有三种情形; (2)只有二个方程有实根,有三种情形; (3)二个方程都有实数根,有一种情形。 从反面,即从否定的角度理解“至少”,只有一种情形:三个方程均无实根。 从正反两方面的“并”的角度审视下,a的范围是 1. Problem Proposal 1 If the following three equations x~2+4ar-4a+3=0, x~2+(a-1)x+a~2=0, x~2+2ax-2a=0 At least one of the equations has a real root and the range of values ​​for the real number a. Analysis: The key word “at least” in the positive understanding of the topic can be obtained as follows: (1) There is only one equation with real roots, and there are three situations; (2) There are only two equations with real roots, and there are three situations; (3) ) Both equations have real roots and there is a situation. To understand “at least” from the negative, that is, from a negative perspective, there is only one case: the three equations have no real roots. From the perspective of both positive and negative “harmony”, the scope of a is