贵刊第七期上刊登了1994年初二《祖冲之杯》数学邀请赛试题。笔者在指导学生解答试题的过程中发现选择题的第一小题条件不足。现求教于同仁。 原题 已知、(2ab)~(1/2)=-2b,其中a、6是有理数。那么((a-b)~2-a)~(1/2)等于( ) (A)-3b;(B)2a-b; In the seventh issue of your magazine, you published the questions for the “Zu Chong Cup” Mathematics Invitational Tournament of the second year of 1994. In the process of guiding students to answer the questions, the author found that the first problem of the multiple-choice questions is insufficient. I am now seeking advice from my colleagues. The original question is known, (2ab)~(1/2)=-2b, where a and 6 are rational numbers. Then ((a-b)~2-a)~(1/2) is equal to () (A)-3b; (B)2a-b;