例1设A=狖x｜x2+4x=0狚,B=狖x｜x2+2(a+1)x+a2-1=0狚.设A∩B=B,求实数a的值.错解由A∩B=B知AB,而A=狖0,-4狚,故0B,有a=±1,-4B.∴a=1或a=7.∴a=±1或a=7.分析错解求出a的值后,没有检验是否符合题意,且没有考虑到B=也是AB的一种情况.应分类讨论:若B≠,求出并验证a的值:(1)当 Example 1 Let A=狖x|x2+4x=0狚,B=狖x|x2+2(a+1)x+a2-1=0狚. Let A∩B=B, find the value of real number a. The wrong solution is known by A∩B=B AB, and A=狖0,-4狚, so 0B, there is a=±1,-4B.∴a=1 or a=7.∴a=± 1 or a=7. Analyze the mistaken solution After finding the value of a, there is no test whether it accords with the meaning of the question, and does not consider that B = is also a kind of AB. Should be classified and discussed: If B≠, find and verify a Value: (1) When