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1．审题不清例1 命题p:(?)∈{(?)};命题q:若A= {1,2},B={x|x(?)A),则A∈B,那么( ) (A)p假,q假． (B)p真,q假． (C)p假,q真． (D)p真,q真．分析命题p显然是真命题,命题q多被学生看成假命题．理由是集合与集合的关系不应是属于,其关键是审题不清,事实上,B= {(?),{1},{2},{1,2}}．这样不难得出A∈B也是真命题,应选(D)．
1. Question 1 is not clear Example 1 Proposition p:(?)∈{(?)}; Proposition q: If A={1,2}, B={x|x(?)A), then A∈B, then () (A) p false, q false. (B)p true, q false. (C)p false, q true. (D)p true, q true. The analysis proposition p is obviously a true proposition, and the proposition q is often seen as a false proposition by students. The reason is that the relationship between collections and collections should not belong, and the key issue is unclear, in fact, B = {(?), {1}, {2}, {1, 2}}. It is not difficult to conclude that A∈B is also a true proposition and should be chosen (D).