论文部分内容阅读
高中《代数》第二册第88页例3告诉我们,重要不等式(χ+y)/2≥(χy)~(1/2)(χ、y∈R~+)可以用来求和的最小值与积的最大值。其意用中文表达出来就是:①两正数积一定,当这两正数相等时,其和最小。②两正数和一定,当两正数相等时,其积最大。“两正数”、“和(或积)一定”、“两正数相等”这三条是一个整体,缺一不可。对此没有足够的认识,在运用重要不等式求最值时,往往出现多种错误。 1 忽视两数是正数例1 求函数y=χ+(1/2)的最值。错解 y=χ+(1/χ)≥2(χ·(1/χ))~(1/2)=2。即y_(最小)=2。分析以上解法未对χ的范围进行讨论,片面地作出结论,忽视了“两正数”的条件。纠正χ>0时,y=x+(1/2)≥2 (χ=1时取等号); χ<0时,-χ>0,-1/χ>0,
The third example of high school Algebra Volume Two, page 88, tells us that the important inequality (χ+y)/2≥(χy)~(1/2)(χ,y∈R~+) can be used to sum the minimum The maximum value and product. Its intention in Chinese is: 1 The positive number is positive, and when the positive numbers are equal, the sum is smallest. Two positive numbers and some, when the two positive numbers are equal, the product is the largest. The three “positive numbers,” “and (or product) certain,” and “two positive numbers are equal” are integral and indispensable. There is not enough understanding of this, and when using the important inequality to find the best value, there are often multiple mistakes. 1 Ignore the case where the two numbers are positive. Find the value of the function y=χ+(1/2). The wrong solution y=χ+(1/χ)≥2(χ(1/χ))~(1/2)=2. That is, y_(min)=2. Analysis of the above solution did not discuss the scope of the embarrassment and concluded it one-sidedly, ignoring the “two positive numbers” condition. When correcting χ>0, y=x+(1/2)≥2 (When χ=1, the equal sign is used); χ<0, -χ>0,-1/χ>0,