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有些竞赛题条件隐蔽,辗转莫测,如按一般的思路,列式繁锁。若能根据倒数的意义,求一种量是另一种量的倒数的方法,不但独具一格,而且简便、易懂。下举几例说明之。例1.有甲乙两个数,如果甲数的小数点向左移两位,就是乙数的1/8,那么甲数是乙数的几倍? 倒数法解:从题中“甲数的小数或向左移两位”译成“甲数的1/100”,现设甲数×1/100=乙数×1/8=1,
Some of the competition title hidden conditions, unpredictable, such as according to the general train of thought. According to the significance of the countdown, find a quantity is the inverse of another quantity method, not only unique, but also simple and easy to understand. Give a few examples to explain. Example 1. There are two numbers A and B, if a number of decimal places to the left two, that is, the number of B 1/8, then the number is a few times the number of B? Countdown solution: From the title “A few decimal places Or to the left two ”translated into“ a few hundredths of a ”, now set a number × 1/100 = B × 1/8 = 1,