《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 1.1.1 不等式的基本性質(zhì)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) 第一章 不等式的基本性質(zhì)和證明的基本方法 1.1.1 不等式的基本性質(zhì)課件 新人教B版選修45（28頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、第一章 不等式的基本性質(zhì)和證明的基本方法1 1.1 1不等式的基本性質(zhì)和一元二次不等不等式的基本性質(zhì)和一元二次不等式的解法式的解法1 1.1 1.1 1不等式的基本性質(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

2、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.掌握比較兩個實數(shù)大小的方法.2.理解不等式的性質(zhì),能運用不等式的性質(zhì)比較大小.3.能運用不等式的性質(zhì)證明不等式等簡單問題.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHO

3、NGNANJUJIAO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.實數(shù)的大小與實數(shù)的運算性質(zhì)之間的關(guān)系設(shè)a,b為兩個實數(shù),它們在數(shù)軸上的點分別記為A,B,如果A落在B的右邊,則稱a大于b,

4、記為ab;如果A落在B的左邊,則稱a小于b,記為ab,a=b,aba-b0;a=ba-b=0;aba-b0,即(x2-x)-(x-2)0.所以x2-xx-2.答案:x2-xx-2【做一做1-2】 設(shè)x=a2b2+5,y=2ab-a2-4a,若xy,則實數(shù)a,b應(yīng)滿足的條件為.解析:xy,x-y=a2b2+5-2ab+a2+4a=(ab-1)2+(a+2)20.ab-10或a+20,即ab1或a-2.答案:ab1或a-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S

5、UITANGLIANXI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2.不等式的基本性質(zhì) 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨

6、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ZHISHIS

7、HULI知識梳理目標(biāo)導(dǎo)航歸納總結(jié)(1)對于性質(zhì)(4)可以看成:若c0,則abacbc;若cbacb,則下列不等式成立的是() 解析:對于選項A,還需有ab0這個前提條件;對于選項B,當(dāng)a,b都為負(fù)數(shù)時不成立,或一正一負(fù)時可能不成立,如2-3,但22(-3)2不正答案: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SUITANG

8、LIANXI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2-2】 下列命題中正確的有.若ab,則ac2bc2;答案: 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI

9、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.使用不等式的性質(zhì)時要注意哪些問題?剖析:(1)在應(yīng)用傳遞性時,如果兩個不等

10、式中有一個帶等號,而另一個不帶等號,那么等號是不能傳遞的.如ab,bcabac2bc2;若無c0這個條件,則abac2bc2就錯了,因為當(dāng)c=0時,取等號.(3)ab0anbn0成立的條件是“n為正整數(shù),且n2”.如果去掉這個條件,取n=-1,a=3,b=2,那么就會出現(xiàn)3-12-1,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

11、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2.比較兩數(shù)(式)大小的常用方法有哪些?它們有什么區(qū)別?剖析: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

12、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)航目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練

13、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ZHISHISHUL

14、I知識梳理目標(biāo)導(dǎo)航題型一題型二題型三題型四作差比較法 分析:直接作差比較需將 展開,過程較為復(fù)雜,式子冗長,可以考慮兩個式子的特點,根據(jù)兩個式子的特點,先把式子變形后,再作差比較大小.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

15、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反思當(dāng)直接作差不容易判斷兩式的大小或者運算量較大時,可觀察式子自身的特點,先變形,再去作差,然后比較大小.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

16、NXI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)航題型一題型二題型四題型三不等式的性質(zhì)【例2】 判斷下列命題的真假,并簡述理由.(1)ab,cda-cb-d

17、;分析:要判斷上述命題的真假,依據(jù)就是實數(shù)的基本性質(zhì)及實數(shù)運算的符號法則,以及不等式的基本性質(zhì),經(jīng)過合理的邏輯推理即可判斷.也可令式中字母取一些特殊值,以檢驗不等式是否成立.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DI

18、ANLITOUXI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四題型三解:(1)假命題.理由:令a=5,b=4,c=3,d=1,有ab,cd,但a-cb0|a|nbn,但|a|n與an可能相等,也可能互為相反數(shù),故(4)為假命題,如a=-2,b=1,n=3時,|a|b0,但a3=-8b0,cd0,e0.證明:cd-d0.ab0,a-cb-d0. (*)目標(biāo)導(dǎo)航DIANLITOUX

19、I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ZHONGN

20、ANJUJIAO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(1)證明不等式的常用方法:直接利用不等式的性質(zhì).最常用的性質(zhì)有傳遞性、可乘性、同向可加性等;作差法或作商法;函數(shù)的單調(diào)性.(2)在直接利用不等式的性質(zhì)證明時,特別注意以下幾點:是否是同向不等式;此性質(zhì)是否可以逆用.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

21、ITOUXI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)航題型一題型二題型三題型四易錯辨析易錯點:由于多次應(yīng)用同向不等式相加(乘)法則導(dǎo)致變量的取值范圍擴大.【例4】 已知f(x)=mx2-n,且-4f(1)-1,-1f(2)5,求f(3)的取值范圍.加減消元,

22、得0m3,1n7,從而,得-7f(3)=9m-n26,即f(3)的取值范圍是-7,26.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ZHONGNANJUJIA

23、O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錯因分析:m,n是兩個相互關(guān)系、相互制約的量,由條件中的不等式通過加減消元在得出0m3,1n7后,并不意味著m,n可以取得0,3及1,7上的一切值.如當(dāng)m=0,n=7時,m-n=-7已不滿足-4m-n-1.此類題一般是先運用待定系數(shù)法把f(3)用f(1),f(2)表示出來,再利用不等式的性質(zhì)求f(3)的范圍.切勿像錯解那樣先求出m,n的范圍,再求f(3)的范圍,這樣會造成變量的取值范圍擴大.目標(biāo)

24、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SUITANGLIAN

25、XI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ZH

26、ONGNANJUJIAO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 51若a0,-1babab2B.ab2abaC.abaab2D.abab2a解析:a0,-1b0,bb2ab2a.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ZHISHISH

27、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 2 3 4 52若ab,則下列不等式中一定成立的是() 答案:C 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重

28、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3已知a,b

29、,c均為實數(shù),下面四個命題中正確命題的個數(shù)是()A.0B.1C.2D.31 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ZHONGNA

30、NJUJIAO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 54實數(shù)a,b,c,d滿足三個條件:dc;a+b=c+d;a+dc,知bdca.答案:bdca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使不等式a2b lg(a-b)0,2a2b+1都成立的a與b的關(guān)系是.解析:由條件可知,a與b同時滿足|a|b| a-b1,ab+1.故有ab+1,且b0.答案:ab+1,且b0