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 2.1.1 平面上的柯西不等式的代數(shù)和向量形式課件 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 2.1.1 平面上的柯西不等式的代數(shù)和向量形式課件 新人教B版選修45》由會員分享，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 2.1.1 平面上的柯西不等式的代數(shù)和向量形式課件 新人教B版選修45（21頁珍藏版）》請?jiān)谘b配圖網(wǎng)上搜索。

1、第二章 柯西不等式與排序不等式及其應(yīng)用2 2.1 1柯西不等式柯西不等式2 2.1 1.1 1平面上的柯西不等式的代數(shù)和向量形式目標(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知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIA

2、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1.認(rèn)識二維形式的柯西不等式.2.理解二維形式的柯西不等式的幾何意義.3.會利用二維形式的柯西不等式進(jìn)行簡單問題的證明.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISH

3、ULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1.二維形式的柯西不等式定理1(柯西不等式的代數(shù)形式)設(shè)a1,a2,b1,b2均為實(shí)數(shù),則定理2(柯西不等式的向量形式)設(shè),為平面上的兩個(gè)向量,則|.當(dāng)及為非零向量時(shí),上式

4、中等號成立向量與共線(平行)存在實(shí)數(shù)0,使得=.當(dāng)或?yàn)榱阆蛄繒r(shí),規(guī)定零向量和任何向量平行,如,中至少有一個(gè)是零向量,則規(guī)定=0,上面的結(jié)果仍然正確.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析

5、SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航【做一做1-1】 若a,bR,且a2+b2=10,則a-b的取值范圍是()解析:(a2+b2)12+(-1)2(a-b)2,a2+b2=10,(a-b)220.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知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGL

6、IANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航【做一做1-2】 已知p,q(0,+),且p3+q3=2,則p+q的最大值為()答案:A 目標(biāo)導(dǎo)航DI

7、ANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂

8、演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航2.定理4(平面三角不等式) 3.定理5(平面三角不等式的向量形式)設(shè),為平面向量,則|-|+|-|-|.當(dāng)-,-為非零向量時(shí),上面不等式中等號成立存在非負(fù)常數(shù),使得-=(-)向量-與-同向,即夾角為零.名師點(diǎn)撥定理4與定理5是同一定理的不同表示形式,它們都可以看作是定理3的變式.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重

9、難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航【做一做2】 已知,為空間向量,求證:|-|+|-|-|.分析:參照平面三角不等式的向量形式的證明方法進(jìn)行證明即可.證明:設(shè)=(a1,a2,a3),=

10、(b1,b2,b3),=(c1,c2,c3),則-=(a1-b1,a2-b2,a3-b3),-=(b1-c1,b2-c2,b3-c3),-=(c1-a1,c2-a2,c3-a3),|-|+|-|-|,且等號成立存在非負(fù)常數(shù),使得-=(-).目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGN

11、ANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航如何理解柯西不等式?剖析:柯西不等式的幾種形式都需要對其結(jié)構(gòu)進(jìn)行理解與記憶,因此,柯西不等式可以理解為有四個(gè)順序的數(shù)來對應(yīng)的一種不等關(guān)系或構(gòu)造成一個(gè)不等式,如基本不等式是由兩個(gè)數(shù)來構(gòu)造,但怎樣構(gòu)造要仔細(xì)體會.例如:結(jié)合具體問題靈活地應(yīng)用柯西不等式. 目標(biāo)導(dǎo)航DIANLI

12、TOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZH

13、ONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四利用柯西不等式的代數(shù)形式證明不等式【例1】 設(shè)a,b(0,+),且a+b=2.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理目標(biāo)導(dǎo)航DIA

14、NLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四反思利用柯西不等式證明某些不等式時(shí),有時(shí)需要將數(shù)學(xué)表達(dá)式適當(dāng)?shù)刈冃?這種變形往往要求具有很高的技巧,必須善于分析題目的特征,根據(jù)題設(shè)條件,綜合地利用添、拆、分解、組合、配方、變量代換、數(shù)形結(jié)合等方法才能發(fā)現(xiàn)問題的本質(zhì),找到突破口.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGN

15、ANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目

16、標(biāo)導(dǎo)航題型一題型二題型四題型三利用柯西不等式的向量形式證明不等式【例2】 已知a,b(0,+),且a+b=1.求證:(ax+by)2ax2+by2.分析:利用柯西不等式的向量形式.當(dāng)且僅當(dāng)存在實(shí)數(shù)0,使得m=n時(shí)等號成立.故(ax+by)2ax2+by2.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂

17、演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四利用柯西不等式解決實(shí)際問題【例3】 在半徑為R的圓內(nèi),求周長最大的內(nèi)接長方形.分析:首先表示出長方形的周長,得出目標(biāo)函數(shù),再利用柯西不等式求解.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNA

18、NJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)

19、導(dǎo)航題型一題型二題型三題型四反思當(dāng)函數(shù)的解析式中含有根號時(shí)常利用柯西不等式求解. 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

20、聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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):應(yīng)用柯西不等式時(shí),因不注重等號是否成立,從而導(dǎo)致結(jié)果錯(cuò)誤.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGL

21、IANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 5A.a2+b2B.2abC.(a+b)2D.4ab答案:C 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITO

22、UXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 52已知a+b=1,則a2+b2的最小值為() 答案:C

23、 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGL

24、IANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 53若3x+4y=2,則x2+y2的最小值為() 答案:D 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理目標(biāo)導(dǎo)航DIANL

25、ITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 5目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(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知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航1 2 3 4 55設(shè)實(shí)數(shù)x,y滿足3x2+2y26,則p=2x+y的最大值是.