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有这样一道小学数学竞赛辅导训练题:“有多少对正整数x、y(x≤y)满足(1/x)+(1/y)=(1/1992)?” 这是一个不定方程,通常都有多组解,小学生解答起来较为困难。下面介绍一种算术解法。 这道题换句话说就是:(1/1992)可以分成多少对单位分数(分子为1的最简分数)的和。下面我们来讨论把一个单位分数分成两个单位分数的和的计算方法。 先看一个简单的例子:在下面的括号里填上适当的数。
There is such a primary school mathematics competition counseling training questions: “how many pairs of positive integer x, y (x ≤ y) to meet (1 / x) + (1 / y) = (1/1992)?” This is an indefinite equation, There are usually many sets of solutions, and the pupils are more difficult to answer. Here's an arithmetic solution. In other words, this question is: (1/1992) The sum of how many pairs of unit scores (the simplest fraction of the numerator is 1). Let us discuss the calculation method of dividing a unit score into the sum of two unit scores. Take a look at a simple example: Type the appropriate number in the brackets below.