《高中數(shù)學 第一章 不等式的基本性質(zhì)和證明的基本方法 1.5.3 反證法和放縮法課件 新人教B版選修45》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第一章 不等式的基本性質(zhì)和證明的基本方法 1.5.3 反證法和放縮法課件 新人教B版選修45（23頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、1 1.5 5.3 3反證法和放縮法目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLI

2、TOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1.理解反證法在證明不等式中的應用,掌握用反證法證明不等式的方法.2.掌握放縮法證明不等式的原理,并會用其證明不等式.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHON

3、GNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航1.反證法假設要證明的命題是不正確的,然后利用公理,已有的定義、定理,命題的條件逐步分析,得到和命題的條件(或已證明過的定理,或明顯成立的事實)矛盾的結(jié)論,從而得出原來結(jié)論是正確的,這種方法稱作反證法.名師點撥用反證法證明不等式必須把握以下幾點:(1)必須否定結(jié)論,

4、即肯定結(jié)論的反面,當結(jié)論的反面呈現(xiàn)多樣性時,必須羅列出各種情況,缺少任何一種可能,反證法都是不完全的.(2)反證法必須從否定結(jié)論進行推理,即應把結(jié)論的反面作為條件,且必須根據(jù)這一條件進行推證.否則,僅否定結(jié)論,不從結(jié)論的反面出發(fā)進行推理,就不是反證法.(3)推導出的矛盾可能多種多樣,有的與已知矛盾,有的與假設矛盾,有的與已知事實相違背,推導出的矛盾必須是明顯的.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIA

5、O重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航【做一做1-1】 應用反證法推出矛盾的推導過程中要把下列哪些作為條件使用()結(jié)論相反的判斷,即假設;原命題的條件;公理、定理、定義等;原結(jié)論.A.

6、B.C.D.答案:C【做一做1-2】 實數(shù)a,b,c不全為0的等價條件為()A.a,b,c均不為0B.a,b,c中至多有一個為0C.a,b,c中至少有一個為0D.a,b,c中至少有一個不為0答案:D目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHU

7、LI知識梳理目標導航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知識梳理目標導航2.放縮法在證明不等式時,有時需要將所需證明的不等式的值適當放大(或縮小),使它由繁化簡,達到證明目的,這種方法稱為放縮法.其關鍵在于放大(縮小)要適當.名師點撥用放縮法證明不等式時,常見的放縮依據(jù)或技巧是不等式的傳遞性.縮小分母、擴大分子,分式的值增大;縮小分子、擴大分母,分式的值減小;每一次縮小其和變小,但

8、需大于所求;每一次擴大其和變大,但需小于所求,即不能放縮不夠或放縮過頭,同時放縮有時需便于求和.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJ

9、UJIAO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知識梳理目標導航A.M=1B.M1D.M與1的大小關系不確定解析:分母全換成210,共有210個單項.答案:B【做一做2-2】 lg 9lg 11與1的大小關系是.答案:lg 9lg 111 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO

10、重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航1.反證法中的數(shù)學語言是什么?剖析:反證法適宜證明“存在性問題,唯一性問題”,帶有“至少有一個”或“至多有一個”等字樣的問題,或者說“正難則反”,直

11、接證明有困難時,常采用反證法.下面我們列舉一下常見的涉及反證法的文字語言及其相對應的否定假設:對某些數(shù)學語言的否定假設要準確,以免造成原則性的錯誤,有時在使用反證法時,對假設的否定也可以舉一定的特例來說明矛盾,尤其在一些選擇題中,更是如此.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGN

12、ANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航2.放縮法的尺度把握等問題有哪些?剖析:(1)放縮法的理論依據(jù)主要有:不等式的傳遞性;等量加不等量為不等量;同分子(分母)異分母(分子)的兩個分式大小的比較;基本不等式與絕對值不等式的基本性質(zhì);三角函數(shù)的有界性等.(2)放縮法使用的主要方法:放縮法是不等式證明中最重要的

13、變形方法之一,放縮必須有目標,而且要恰到好處,目標往往要從證明的結(jié)論考查.常用的放縮方法有增項、減項、利用分式的性質(zhì)、利用不等式的性質(zhì)、利用已知不等式、利用函數(shù)的性質(zhì)進行放縮等.比如,目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目

14、標導航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知識梳理目標導航目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITAN

15、GLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航題型一題型二題型三題型四用反證法證明否定性結(jié)論命題 分析:“不能同時”包含情況較多,而其否定“同時大于”僅有一種情況,因此適宜用反證法證明.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJI

16、AO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航題型一

17、題型二題型三題型四目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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

18、SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三題型四反思(1)當證明的結(jié)論中含有“不是”“不都”“不存在”等詞語時,適于應用反證法,因為此類問題的反面比較具體.(2)用反證法證明不等式時,推出的矛盾有三種表現(xiàn)形式:與已知矛盾;與假設矛盾;與顯然成立的事實相矛盾.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHI

19、SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航題型一題型二題型四題型三用反證法證明“至多”“至少”類問題 分析:問題從正面證明不易入手,適合應用反證法證明.目標導航DIANLITOUXI典例透析SUITANGLIA

20、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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ZHI

21、SHISHULI知識梳理目標導航題型一題型二題型四題型三假設不成立, 則|f(1)|+2|f(2)|+|f(3)|a2+ab+b2=a+b,故a+b1.因為(a+b)24ab,目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航D

22、IANLITOUXI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知識梳理目標導航題型一題型二題型三題型四反思用放縮法證明不等式的過程中,往往采用添項或減項的“添舍”放縮,拆項對比的分項放縮,函數(shù)的單調(diào)性放縮等.放縮時要注意適度,否則不能同向傳遞.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI

23、典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航題型一題型二題型三題型四易錯辨析易錯點:在證明不等式時,因不按不等式的性質(zhì)變形

24、,從而導致證明過程錯誤.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典

25、例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 4 51用反證法證明:若整系數(shù)一元二次方程ax2+bx+c=0(a0)有有理根,那么a,b,c中至少有一個偶數(shù).下列假設中正確的是()A.假設a,b,c都是偶數(shù)B.假設a,b,c都不是偶數(shù)C.假設a,b,c中至多有一個偶數(shù)D.假設a,b,c中至多有兩個偶數(shù)答案:B目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練

26、ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航1 2 3 4 52設x,y(0,+),且xy-(x+1)=1,則() 答案:B 目標導航DIANLITOUXI典例透析

27、SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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ZHONGNANJUJ

28、IAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 4 5A.都大于2B.都小于2C.至少有一個不大于2D.至少有一個不小于2答案:D 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析

29、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知識梳理目標導航1 2 3 4 5答案: 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航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知識梳理目標導航1 2 3 4 55若正數(shù)a,b滿足ab1+a+b,則a+b的最小值為.