高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)用 2.3 平均值不等式（選學(xué)）課件 新人教B版選修45

《高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)用 2.3 平均值不等式（選學(xué)）課件 新人教B版選修45》由會(huì)員分享，可在線閱讀，更多相關(guān)《高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)用 2.3 平均值不等式（選學(xué)）課件 新人教B版選修45（20頁珍藏版）》請?jiān)谘b配圖網(wǎng)上搜索。

1、2 2.3 3平均值不等式平均值不等式(選學(xué)選學(xué))目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳

2、理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1.了解算術(shù)平均,幾何平均,調(diào)和平均的概念.2.理解定理的意義及作用,了解定理的推證過程.3.能夠靈活應(yīng)用定理證明求解一些簡單問題.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANG

3、LIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1.有關(guān)概念 2.定理定理1(算術(shù)幾何平均值不等式,簡稱平均值不等式)(2)推論1:設(shè)a1,a2,an為n個(gè)正數(shù),且a1a2an=1,則a1+a2+ann,且等號(hào)成立a1=a2=an=1.(3)推論2:設(shè)C為常數(shù),且a1,a2,an為n個(gè)

4、正數(shù),則當(dāng)a1+a2+an=nC時(shí),a1a2anCn,且等號(hào)成立a1=a2=an.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

5、聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航定理2 定理3 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳

6、理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航加權(quán)平均不等式 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI

7、典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航【做一做1】 下列命題是假命題的是() 答案:A 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUX

8、I典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航【做一做2】 已知x,y,z(0,+),且2x+3y+5z=6,則xyz的最

9、大值為.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITA

10、NGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航平均值不等式的應(yīng)用條件是什么?剖析:“一正”:不論是三個(gè)數(shù)的或者n個(gè)數(shù)的平均值不等式,都要求這三個(gè)數(shù)或者n個(gè)數(shù)都是正數(shù),否則不等式是不成立的;“二定”:包含兩類求最值問題,一是已知n個(gè)正數(shù)的和為定值(即a1+a2+an為定值),求其積a1a2an的最大值,二是已知乘積a1a2an為定值,求其和a1+a2+an的最小值;“三相等”:取等號(hào)的條件是a1=a2=an,不能只有其中一部分相等.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHI

11、SHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四

12、利用平均值不等式證明不等式 分析:觀察求證式子的結(jié)構(gòu),通過變形轉(zhuǎn)化為用平均值不等式證明.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIA

13、O重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四反思不等式的證明方法較多,關(guān)鍵是從式子的結(jié)構(gòu)入手進(jìn)行分析,找到證明不等式的突破口,使其出現(xiàn)平均值不等式的形式.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DI

14、ANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型四題型三利用平均值不等式求最值 分析:對(duì)于x2(1-5x),視x2與1-5x為兩項(xiàng),其和不可能為定值,應(yīng)把x2拆為兩項(xiàng)x,x,故x,x,(1-5x)這三項(xiàng)同時(shí)配系數(shù)才能使和為定值.目

15、標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIA

16、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型四題型三反思本題采用的方法是拆項(xiàng),把x2變?yōu)閤,x,再配系數(shù)的方法,請思考采用下面的變形錯(cuò)在什么地方?目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISH

17、ISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四平均值不等式的應(yīng)用【例3】 某同學(xué)在電腦城組裝了一臺(tái)電腦,總費(fèi)用為3 600元.假定在電腦的使用過程中,維修費(fèi)平均為:第一年200元,第二年400元,第三年600元,依等差數(shù)列逐年遞增,問:這臺(tái)電腦使用多少年報(bào)廢最合算?分析:要求電腦使用多少年報(bào)廢最合算,實(shí)際上是求使用多少年的平

18、均費(fèi)用最少,這種年平均費(fèi)用一般由兩部分組成:一部分是電腦成本的平均值,另一部分是電腦維修費(fèi)用的平均值.這樣,電腦的最值報(bào)廢年限,即:年平均消耗費(fèi)用=年均成本費(fèi)+年均維修費(fèi),電腦最佳報(bào)廢年限=年均消耗費(fèi)用最低的年限.因此,需建立年平均費(fèi)用y與使用年數(shù)x的函數(shù)關(guān)系式.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIAN

19、XI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四所以當(dāng)x=6(年)時(shí),y有最小值,即這臺(tái)電腦使用6年報(bào)廢最合算.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLI

20、TOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四易錯(cuò)辨析易錯(cuò)點(diǎn):忽視應(yīng)用平均值不等式求最值應(yīng)

21、具備的“一正、二定、三相等”而致錯(cuò).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLI

22、TOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透

23、析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 41若x,yR,且xy0,x2y=2,則xy+x2的最小值是.答案:3 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理

24、目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 4目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例

25、透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 4目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂

26、演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 44求證:在表面積一定的長方體中,以正方體的體積最大.證明:設(shè)長方體的三條相交于同一頂點(diǎn)的棱長分別為x,y,z,則長方體的體積為V=xyz,表面積A=2xy+2yz+2zx,根據(jù)平均值不等式,