高中數(shù)學(xué) 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

《高中數(shù)學(xué) 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(huì)員分享，可在線(xiàn)閱讀，更多相關(guān)《高中數(shù)學(xué) 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(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、1 1.5 5.3 3反證法和放縮法目標(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

2、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1.理解反證法在證明不等式中的應(yīng)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(huì)用其證明不等式.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典例透析SUITANGLIANXI隨堂演練ZHON

3、GNANJUJIAO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.反證法假設(shè)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(guò)的定理,或明顯成立的事實(shí))矛盾的結(jié)論,從而得出原來(lái)結(jié)論是正確的,這種方法稱(chēng)作反證法.名師點(diǎn)撥用反證法證明不等式必須把握以下幾點(diǎn):(1)必須否定結(jié)論,

4、即肯定結(jié)論的反面,當(dāng)結(jié)論的反面呈現(xiàn)多樣性時(shí),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(jìn)行推理,即應(yīng)把結(jié)論的反面作為條件,且必須根據(jù)這一條件進(jìn)行推證.否則,僅否定結(jié)論,不從結(jié)論的反面出發(fā)進(jìn)行推理,就不是反證法.(3)推導(dǎo)出的矛盾可能多種多樣,有的與已知矛盾,有的與假設(shè)矛盾,有的與已知事實(shí)相違背,推導(dǎo)出的矛盾必須是明顯的.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

5、O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1-1】 應(yīng)用反證法推出矛盾的推導(dǎo)過(guò)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(shè);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(shí)數(shù)a,b,c不全為0的等價(jià)條件為()A.a,b,c均不為0B.a,b,c中至多有一個(gè)為0C.a,b,c中至少有一個(gè)為0D.a,b,c中至少有一個(gè)不為0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(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ZHISHISHU

7、LI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2.放縮法在證明不等式時(shí),有時(shí)需要將所需證明的不等式的值適當(dāng)放大(或縮小),使它由繁化簡(jiǎn),達(dá)到證明目的,這種方法稱(chēng)為放縮法.其關(guān)鍵在于放大(縮小)要適當(dāng).名師點(diǎn)撥用放縮法證明不等式時(shí),常見(jiàn)的放縮依據(jù)或技巧是不等式的傳遞性.縮小分母、擴(kuò)大分子,分式的值增大;縮小分子、擴(kuò)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(kuò)大其和變大,但需小于所求,即不能放縮不夠或放縮過(guò)頭,同時(shí)放縮有時(shí)需便于求和.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ZHONGNANJ

9、UJIAO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A.M=1B.M1D.M與1的大小關(guān)系不確定解析:分母全換成210,共有210個(gè)單項(xiàng).答案:B【做一做2-2】 lg 9lg 11與1的大小關(guān)系是.答案:lg 9lg 111 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

10、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.反證法中的數(shù)學(xué)語(yǔ)言是什么?剖析:反證法適宜證明“存在性問(wèn)題,唯一性問(wèn)題”,帶有“至少有一個(gè)”或“至多有一個(gè)”等字樣的問(wèn)題,或者說(shuō)“正難則反”,直

11、接證明有困難時(shí),常采用反證法.下面我們列舉一下常見(jiàn)的涉及反證法的文字語(yǔ)言及其相對(duì)應(yīng)的否定假設(shè):對(duì)某些數(shù)學(xué)語(yǔ)言的否定假設(shè)要準(zhǔn)確,以免造成原則性的錯(cuò)誤,有時(shí)在使用反證法時(shí),對(duì)假設(shè)的否定也可以舉一定的特例來(lái)說(shuō)明矛盾,尤其在一些選擇題中,更是如此.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ZHONGN

12、ANJUJIAO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2.放縮法的尺度把握等問(wèn)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(gè)分式大小的比較;基本不等式與絕對(duì)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(biāo),而且要恰到好處,目標(biāo)往往要從證明的結(jié)論考查.常用的放縮方法有增項(xiàng)、減項(xiàng)、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(jìn)行放縮等.比如,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

14、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SUITAN

15、GLIANXI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用反證法證明否定性結(jié)論命題 分析:“不能同時(shí)”包含情況較多,而其否定“同時(shí)大于”僅有一種情況,因此適宜用反證法證明.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJI

16、AO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)航題型一

17、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典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識(shí)梳理DIANLITOUXI典例透析

18、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(1)當(dāng)證明的結(jié)論中含有“不是”“不都”“不存在”等詞語(yǔ)時(shí),適于應(yīng)用反證法,因?yàn)榇祟?lèi)問(wèn)題的反面比較具體.(2)用反證法證明不等式時(shí),推出的矛盾有三種表現(xiàn)形式:與已知矛盾;與假設(shè)矛盾;與顯然成立的事實(shí)相矛盾.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ZHISHI

19、SHULI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四題型三用反證法證明“至多”“至少”類(lèi)問(wèn)題 分析:問(wèn)題從正面證明不易入手,適合應(yīng)用反證法證明.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIA

20、NXI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SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHI

21、SHISHULI知識(shí)梳理目標(biāo)導(dǎo)航題型一題型二題型四題型三假設(shè)不成立, 則|f(1)|+2|f(2)|+|f(3)|a2+ab+b2=a+b,故a+b1.因?yàn)?a+b)24ab,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D

22、IANLITOUXI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反思用放縮法證明不等式的過(guò)程中,往往采用添項(xiàng)或減項(xiàng)的“添舍”放縮,拆項(xiàng)對(duì)比的分項(xiàng)放縮,函數(shù)的單調(diào)性放縮等.放縮時(shí)要注意適度,否則不能同向傳遞.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

23、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):在證明不等式時(shí),因不按不等式的性質(zhì)變形

24、,從而導(dǎo)致證明過(guò)程錯(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(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

25、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 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(gè)偶數(shù).下列假設(shè)中正確的是()A.假設(shè)a,b,c都是偶數(shù)B.假設(shè)a,b,c都不是偶數(shù)C.假設(shè)a,b,c中至多有一個(gè)偶數(shù)D.假設(shè)a,b,c中至多有兩個(gè)偶數(shù)答案:B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

26、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 52設(shè)x,y(0,+),且xy-(x+1)=1,則() 答案:B 目標(biāo)導(dǎo)航DIANLITOUXI典例透析

27、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SUITANGLIANXI隨堂演練ZHONGNANJUJ

28、IAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 4 5A.都大于2B.都小于2C.至少有一個(gè)不大于2D.至少有一個(gè)不小于2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(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

29、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 5答案: 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 55若正數(shù)a,b滿(mǎn)足ab1+a+b,則a+b的最小值為.