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例1 方程|x-2|+|x-3|=1的实数解的个数有( )。 (A)2个 (B)3个 (C)4个 (D)无数多个 (1992年“祖冲之杯”初中数学邀请赛试题) 解易见当x>3或x<2时,无解。而当2≤x≤3时, |x-2|+|x-3|=|x-2|+|3-x|=|x-2+3-x|=1。所以 2≤x≤3,故原方程有无数多个实数解。故答案选(D)。说明此题一般解法是分段讨论,求方程的解。现另辟蹊径,采用|a+b|=|a|+|b|当且仅当ab≥0时成立,能取到意想不到的效果。当然,采用|a+b|=|a|+|b|要特别注意条件。
Example 1 The number of real solutions of the equation |x-2|+|x-3|=1 is (). (A) 2 (B) 3 (C) 4 (D) Innumerable multiple (in 1992 “Zuchong Cup” junior high school mathematics invitational exam questions) The solution is easy to see when x>3 or x<2, no solution. When 2≤x≤3, |x−2|+|x−3|=|x−2|+|3-x|=|x−2+3-x|=1. So 2 ≤ x ≤ 3, so the original equation has countless real solutions. So answer selection (D). Explain that the general solution of this problem is a piecewise discussion and find the solution of the equation. Another way is to use |a+b|=|a|+|b| if and only if ab≥0 holds true for unexpected results. Of course, use |a+b|=|a|+|b| to pay special attention to the conditions.