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在初中数学竞赛中,常出现有关整式求值问题. 例1 设a是最小的正整数,b是最大的负整数,c是绝对值最小的有理数,则a-b+c=( ). (A)-1 (B)0 (C)1 (D)2 (1999年“希望杯”数学邀请赛初一试题) 解由题意知a=1,b=-1,c=0. 原式=1-(-1)+0=2.故选D. 例2 已知2a~2b~(n-1)与-3a~2b~(2)m是同类项,那么(2m-n)~x=__.(第十五届江苏省初中数学竞赛初一试题) 解由同类项定义知x=2,n-1=2m. 所以2m-n=-1.于是(2m-n)~x=(-1)~2=1. 说明正确掌握有理数、同类项等有关概念是解这类题的关键.
In the junior high school math competition, there is often a question about the evaluation of integers. Example 1 Let a be the smallest positive integer, b the largest negative integer, and c the rational number with the smallest absolute value, then a-b+c=( ). A)-1 (B)0 (C)1 (D)2 (The 1999 “Hope Cup” Math Invitational Preliminary Questions) The solution from the question is known as a=1, b=-1, c=0. 1-(-1)+0=2. So choose D. Example 2 It is known that 2a~2b~(n-1) and -3a~2b~(2)m are the same type terms, then (2m-n)~x =__. (The 15th Jiangsu Provincial Junior High School Mathematics Competition Preliminary Questions) The solution is known by the definition of similar items x=2,n-1=2m. So 2m-n=-1. Then (2m-n)~x= (-1)~2=1. It is necessary to correctly grasp the concepts of rational numbers and similar items as the key to solving such problems.