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六年制代数二册课本P_(106)定理一的推论为:|a_1+a_2+…+a_n|≤|a_1|+|a_2|+…+|a_n|(※)此式可据定理一采用数学归纳法来证。但课本上并没有指明(※)中等号成立的条件,试问(※)取等号的条件是什么?易知,当且仅当a_i同为非负或同为非正时,(※)取等号。此结论用来解某些含绝对值的问题对,较之一般使用分区间去绝对值号的方法要简明得多,现分别举例说明如下。例1 求证|x十1/x|≥2(x≠0) 证明∵x≠0,∴x与1/x恒同号,依(※)取等号的条件得|x+1/x|=|x|+|1/x|≥2((|x||1/x|)~(1/2))=2.
The corollary of the six-year algebra two-volume textbook P_(106) Theorem is: |a_1+a_2+...+a_n|≤|a_1|+|a_2|+...+|a_n| (※) This formula can be used according to the theorem Mathematical induction to prove. However, the textbook does not specify the conditions for (※) the establishment of an equal number, what is the condition of the question (※) to take an equal sign, and is easy to know, if and only if a_i is non-negative or non-positive, (※) equal sign. This conclusion can be used to solve certain problem pairs with absolute values, which is much more concise than the method of using absolute value between partitions. Example 1 Verification|x|1/x|≥2(x≠0) Prove that ∵x≠0, ∴x, and 1/x constant symbols, and (※) the condition of the equal sign |x+1/x| =|x|+|1/x|≥2((|x||1/x|)~(1/2))=2.