《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 1.2 基本不等式課件 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) 第一章 不等式的基本性質(zhì)和證明的基本方法 1.2 基本不等式課件 新人教B版選修45（29頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、1 1.2 2基本不等式基本不等式目標(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DIANLIT

2、OUXI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.了解兩個(gè)或三個(gè)正數(shù)的算術(shù)平均值和幾何平均值.2.理解定理1和定理2(基本不等式).3.探索并了解三個(gè)正數(shù)的算術(shù)幾何平均值不等式的證明過程.4.掌握用基本不等式求一些函數(shù)的最值及實(shí)際的應(yīng)用問題.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)

3、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.定理1設(shè)a,bR,則a2+b22ab,當(dāng)且僅當(dāng)a=b時(shí),等號(hào)成立.2.定理2(基本不等式或平均值不等式) (3)基本不等式可用語(yǔ)言敘述為:兩個(gè)正數(shù)的算術(shù)平均值大于或等于它們的幾何平均值.

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SUITANGLI

5、ANXI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ZHONGNANJUJI

6、AO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2-1】 下列不等式中正確的是()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ZHONGNANJU

7、JIAO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)航答案: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典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI

8、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3.定理3(三個(gè)正數(shù)的算術(shù)幾何平均值不等式或平均值不等式)(3)定理3可用語(yǔ)言敘述為三個(gè)正數(shù)的算術(shù)平均值不小于它們的幾何平均值.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITAN

9、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

10、焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航【做一做3】 已知x,y,z是正數(shù),且x+y+z=6,則lg x+lg y+lg z的取值范圍是()A.(-,lg 6B.(-,3lg 2C.lg 6,+)D.3lg 2,+)解析:x,y,z是正數(shù),lg x+lg y+lg z=lg xyzlg 23=3lg 2,當(dāng)且僅當(dāng)x=y=z=2時(shí),等號(hào)成立.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ZHONGNANJUJIAO重難聚焦ZH

11、ISHISHULI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4.定理4(一般形式的算術(shù)幾何平均值不等式) 答案:4 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

12、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1.三個(gè)或三個(gè)

13、以上正數(shù)的算術(shù)幾何平均值不等式的應(yīng)用條件是什么?剖析:“一正”:不論是三個(gè)數(shù)的平均值不等式或者n個(gè)數(shù)的平均值 “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”:等號(hào)成立的條件是a1=a2=a3=an,不能只是其中一部分值相等.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ZHONGNANJUJI

14、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.如何使用基本不等式中的變形與拼湊方法?剖析:為了使用基本不等式求最值(或范圍等),往往需要對(duì)數(shù)學(xué)代數(shù)式變形或拼湊數(shù)學(xué)結(jié)構(gòu),有時(shí)一個(gè)數(shù)拆成兩個(gè)

15、或兩個(gè)以上的數(shù),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S

16、UITANGLIANXI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利用基本不等式比較大小 分析:解答本題應(yīng)充分利用基本不等式及其變形,不等式的性質(zhì).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ZH

17、ISHISHULI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

18、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DIANLIT

19、OUXI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四題型三利用基本不等式求最值 分析:根據(jù)題設(shè)條件,合理

20、變形,創(chuàng)造能用基本不等式的條件. 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DIANLIT

21、OUXI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

22、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SUITANGLIANX

23、I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反思利用基本不等式解題時(shí)要注意考察“三要素”:(1)函數(shù)中的相關(guān)項(xiàng)必須都是正數(shù);(2)變形后各項(xiàng)的和或積有一個(gè)必須是常數(shù);(3)當(dāng)且僅當(dāng)各項(xiàng)相等時(shí),才能取到等號(hào),可簡(jiǎn)化為“一正二定三相等”.求函數(shù)的最值時(shí),常將不滿足上述條件

24、的函數(shù)式進(jìn)行“拆”、“配”等變形,使其滿足條件,進(jìn)而求出最值.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ZHISHISH

25、ULI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基本不等式的實(shí)際應(yīng)用【例3】 某國(guó)際化妝品生產(chǎn)企業(yè)為了占有更多的市場(chǎng)份額,擬在2016年第31屆夏季奧林匹克運(yùn)動(dòng)會(huì)期間進(jìn)行一系列促銷活動(dòng),經(jīng)過市場(chǎng)調(diào)查和測(cè)算,化妝品的年銷量x萬(wàn)件與年促銷費(fèi)t萬(wàn)元之間滿足3-x與t+1成反比例的關(guān)系,如果不搞促銷活動(dòng),化妝品的年銷量只能是1萬(wàn)件,已知2016年生產(chǎn)化妝品的設(shè)備折舊、維修等固定費(fèi)用為3萬(wàn)元,每生產(chǎn)1萬(wàn)件化妝品需要投入32萬(wàn)元的生產(chǎn)費(fèi)用,若將每件化妝品的售價(jià)定為其生產(chǎn)成本的15

26、0%與平均每件促銷費(fèi)的一半之和,則當(dāng)年生產(chǎn)的化妝品正好能銷完.(1)將2016年的利潤(rùn)y(單位:萬(wàn)元)表示為促銷費(fèi)t(單位:萬(wàn)元)的函數(shù);(2)該企業(yè)2016年的促銷費(fèi)投入多少萬(wàn)元時(shí),企業(yè)的年利潤(rùn)最大?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ZHISHIS

27、HULI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分析:表示出題中的所有已知量和未知量,利用它們之間的關(guān)系列出函數(shù)表達(dá)式,再應(yīng)用不等式求最值.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SUITANGL

28、IANXI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目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHO

29、NGNANJUJIAO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)

30、梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四反思解答不等式的實(shí)際應(yīng)用問題,一般可分為如下四步:閱讀理解材料:應(yīng)用題所用語(yǔ)言多為“文字語(yǔ)言、符號(hào)語(yǔ)言、圖形語(yǔ)言”并用,而且多數(shù)應(yīng)用題篇幅較長(zhǎng).閱讀理解材料要達(dá)到的目的是將實(shí)際問題抽象成數(shù)學(xué)模型.這就要求解題者領(lǐng)悟問題的實(shí)際背景,確定問題中量與量之間的關(guān)系,初步形成用怎樣的模型能夠解決問題的思路,明確解題方向.建立數(shù)學(xué)模型:根據(jù)中的分析,把實(shí)際問題用“符號(hào)語(yǔ)言”、“圖形語(yǔ)言”抽象成數(shù)學(xué)模型,并且建立所得數(shù)學(xué)模型和已知數(shù)學(xué)模型的對(duì)應(yīng)關(guān)系,以便確立下一步的努力方向.討論不等關(guān)系:根據(jù)題目要求和中建立起來的數(shù)學(xué)模型,討論與結(jié)論有關(guān)的不等關(guān)系,得出有關(guān)理論參數(shù)的值.

31、得出問題結(jié)論:根據(jù)中得到的理論參數(shù)的值,結(jié)合題目要求得出問題的結(jié)論.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ZHISH

32、ISHULI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):利用基本不等式求最值時(shí),應(yīng)注意不等式成立的條件,即變量為正實(shí)數(shù),和或積為定值,等號(hào)成立,三者缺一不可.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DIANLITO

33、UXI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SUITA

34、NGLIANXI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 51下列函數(shù)中,最小值為2的是() 答案:D 目標(biāo)導(dǎo)航DIANLITOUXI典例

35、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ZHONGNANJ

36、UJIAO重難聚焦ZHISHISHULI知識(shí)梳理目標(biāo)導(dǎo)航1 2 3 4 5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(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

37、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 5A.3B.4C.5D.6 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ZHONGNANJUJI

38、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 54周長(zhǎng)為l的矩形的面積的最大值為,對(duì)角線長(zhǎng)的最小值為. 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