深挖课本例题的潜在功能一例

• 作 者: 郭允远
• 机 构: 山东沂南县教委教研室 (276300)
• 出 处: 考试
• 发表日期: 2000年12期
• 关键词: 潜在功能 构造函数 正分数 证法 高考命题 学生智能 解题能力 通项 复习课 数学归纳法

高中课本《代数》下册(必修本)第12页例7:已知 a,b,m∈R~+,并且 a(a/b).1.探究其它证法本例在课本中是作为分析法证明不等式给出的,用比较法也容易证明.若注意观察不等式两边的结构特点,又可获得构造函数,利用函数单词性证明的新思路.证明:构造函数 f(x)=(a+x)/(b+x),则 f(x)=1+(b+a)/(b+x),∵a0,故函数 f(x)在区间(-b,+∞)上是增函数.由 m>0,得 f(m)>f(O),即(a+m)/(b+m)>(a/b).2.发现正分数的两条性质 High School Textbook Algebra, Second Edition (required) Page 12 Example 7: Known a, b, m∈R~+, and a(a /b).1. Exploring other proofs This example is given in the textbook as an analytical inequality, and it is easy to prove with a comparative method. If you look at the structural features of both sides of the inequality, you can obtain a constructor and use the functional word. New idea of ​​proof of nature. Proof: Constructor f(x)=(a+x)/(b+x), then f(x)=1+(b+a)/(b+x), ∵a< b, ∴ba>0, so the function f(x) is an increasing function on the interval (-b,+∞). From m>0, we get f(m)>f(O), ie (a+m)/ (b+m)>(a/b).2. Two properties of finding positive scores