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关于函数y=|(x-a_1)~2+(b_1~2)~(1/2)±(x-a_2)~2+(b_2~2)~(1/2)的最值问题的解法,很多文章里都有论述,大多是采用求导法,图象法或两点距离公式来研究的。这里介绍一种用复数的模以及有关性质来解的方法。问题1 求函数g(x)=(x-a_1)~2+(b_1~2)~(1/2)+(x-a_2)~2+(b_2~2)~(1/2)的最小值。解因题中条件与b_1、b_2的正、负无关,我们可只考虑b_1、b_2为正值。并设z_1=(x-a_1)
The Solution to the Problem of the Function y=|(x-a_1)~2+(b_1~2)~(1/2)±(x-a_2)~2+(b_2~2)~(1/2) Many articles have been discussed. Most of them are studied using the derivative method, image method or two-point distance formula. Here we introduce a method of solving with a complex modulus and related properties. Problem 1 Find the minimum of the function g(x)=(x-a_1)~2+(b_1~2)~(1/2)+(x-a_2)~2+(b_2~2)~(1/2) value. The condition of solution problem is independent of the positive and negative of b_1 and b_2. We can only consider b_1 and b_2 as positive values. And set z_1=(x-a_1)