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对于一些初看有难度不易解决的问题,我们可以在不改变它的本义的情况下,通过变换它的形式,让它变得通俗易懂,从而找到解题的思路。【例】已知11/a+11/b+11/c=143/210,且a≠b≠c。求a、b、c各是多少。【思路点睛】在计算异分母分数加减法时,常常需要先通分再计算。题中三个分数的分子都是11,通分后三个加数的分子的和一定是11的倍数,又因为143=11×13,所以原题可变换
For some problems that are difficult to solve at first glance, we can find a solution to the problem by changing its form and making it easy to understand without changing its original meaning. [Example] It is known that 11 / a + 11 / b + 11 / c = 143/210 and a ≠ b ≠ c. Seek a, b, c is how much each. 【Ideas dotting】 In the calculation of different denominator fraction addition and subtraction method, you often need to re-calculate the first pass. Problem three molecules are 11, through the sum of three molecules after the sum must be a multiple of 11, and because 143 = 11 × 13, so the original title can be transformed