1 1.5 5.2 2综合法和分析法目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1.理解综合法、分析法证明不等式的原理和思想.2.掌握综合法、分析法证明简单不等式的方法和步骤.3.能综合运用综合法、分析法证明不等式.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1.综合法在证明不等式的时候,我们经常要从命题的已知条件出发,利用公理、已知的定义及定理,逐步推导,从而最后导出要证明的命题,这种方法称为综合法.名师点拨用综合法证明不等式,就是用因果关系书写“从已知出发,借助不等式的性质和有关定理,经过逐步的逻辑推理,最后达到待证不等式得证”的全过程,其特点可描述为“由因导果”,即从“已知”看“可知”,逐步推向“未知”.综合法属逻辑方法范畴,它的严谨体现在步步注明推理依据.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航答案:D 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航2.分析法从需要证明的命题出发,分析使这个命题成立的充分条件,利用已知的一些定理,逐步探索,最后达到命题所给出的条件(或者一个已证明过的定理或一个明显的事实),这种证明方法称为分析法.归纳总结目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航名师点拨用分析法证明“若A则B”的模式为:欲证命题B成立,只需证命题B1成立,只需证命题B2成立,只需证明A为真.已知A为真,故B必为真.可以简单写成:BB1B2BnA.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航【做一做2】要证a2+b2-1-a2b20,只要证() 解析:要证a2+b2-1-a2b20,只要证a2(1-b2)+(b2-1)0,只要证(b2-1)(a2-1)0.答案:D 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航综合法在应用中的有关问题是什么?剖析:用综合法证明不等式时,主要利用基本不等式,函数的单调性以及不等式的性质,在严密的演绎推理下推导出结论.综合法证明问题的“入手处”是题设中的已知条件或某些基本不等式.比如下面的几个,是经常用到的:目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航选择使用哪个不等式作为证题的“原始出发点”或对已知条件的转化是证题的关键,这要求对要证明的结果有充分的分析过程,可以联系平时学习过程中积累下来的数学结论或知识作出判断.比如目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三用综合法证明不等式【例1】设a,b,c为ABC的三条边,求证:a2+b2+c22ab+2bc+2ac.分析:本题看似是一道与公式a2+b22ab(a,bR)有关的题目,又好像与二次函数有关,但实际上这两种思路都行不通.其实本题的关键在于ABC中隐含的a,b,c的关系.证明:证法一:在ABC中,ab+c,ba+c,cb+a,则a2a(b+c),b2b(a+c),c2c(b+a).于是a2+b2+c2a(b+c)+b(a+c)+c(a+b),即a2+b2+c2bc,则0a-bc,0b-ca,0a-cb.(a-b)2c2,(b-c)2a2,(a-c)2b2.(a-b)2+(b-c)2+(a-c)2c2+a2+b2,即a2+b2+c22ab+2bc+2ac.证法三:在ABC中,设a,b,c三边所对的角分别为角A,B,C,则由余弦定理,得a2=b2+c2-2bccos A,b2=a2+c2-2accos B,c2=a2+b2-2abcos C,则a2+b2+c2=(b2+c2-2bccos A)+(a2+c2-2accos B)+(a2+b2-2abcos C),即a2+b2+c2=2abcos C+2bccos A+2accos B.在ABC中,cos A1,cos B1,cos C1,2abcos C+2bccos A+2accos B2ab+2bc+2ca.a2+b2+c20.则f(a)的值可正、可负、可为零,无法确定.因此,分析题目时,对条件要看清楚,尤其要探寻条件间的限制关系,以免受到某些思维定式的影响.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三用分析法证明不等式 分析:本题要证明的不等式显得较为复杂,不易观察出怎样由ab0得到要证明的不等式,因而可以用分析法先变形要证明的不等式,从中找到证题的线索.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三反思分析法的格式是固定化的,其中每一步都是上一步成立的充分条件,即每一步数学式的变化都是在这个要求之下一步一步去寻找成立的条件或结论、定理.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三易错辨析易错点:因不注意分析法的证题格式而致错.错因分析:分析法是“执果索因”,即由待证的不等式出发,逐步寻求它要成立的充分条件,所以书写时必须按照分析法的固有格式进行.本题的错误就在于未按分析法的格式证明.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5A.AB B.ABC.ABD.Ay1,0aayB.logaxlogayC.xayaD.x-ay-a解析:0ay1,根据指数函数、对数函数、幂函数的单调性可知axay,logaxya,故排除选项A,B,C.答案:D 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5A.2B.3C.4D.5答案:C 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5答案:a+b 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 5答案: