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在数学竞赛中,经常遇到各类数值计算题,这类题一般数字冗繁,如果不运用合理的计算方法是很难获得正确结果的.现举例如下. 例1 求值 201224/20002~2-1999·2001. 解∵1999·2001=(2000-1)(2000+1)=2000~2-1, ∴201224/20002~2-1999·2001=201224. 例2 求值1995~3-2×1995~2-1993/1995~3+1995~2-1996 (1995年北京市数学竞赛题) 解原式=1995~2(1995-2)-(1995-2)/1995~2(1995+1)-(1995+1) =(1995-2)(1995~2-1)/(1995+1)(1995~2-1) =1993/1996
In mathematics competitions, various types of numerical calculation problems are often encountered. These types of questions are generally tedious in number, and it is difficult to obtain correct results without using a reasonable calculation method. Examples are given below. Example 1 Evaluation 201224/20002~2- 1999·2001. Solution ∵1999·2001=(2000-1)(2000+1)=2000~2-1, ∴201224/20002~2-1999·2001=201224. Example 2 Evaluation 1995~3-2× 1995~2-1993/1995~3+1995~2-1996 (Beijing Mathematical Competition in 1995) Solution of the original formula = 1995~2(1995-2)-(1995-2)/1995~2(1995+1 )-(1995+1)=(1995-2)(1995~2-1)/(1995+1)(1995~2-1)=1993/1996