第一章 不等式的基本性质和证明的基本方法1 1.1 1不等式的基本性质和一元二次不等不等式的基本性质和一元二次不等式的解法式的解法1 1.1 1.1 1不等式的基本性质目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1.掌握比较两个实数大小的方法.2.理解不等式的性质,能运用不等式的性质比较大小.3.能运用不等式的性质证明不等式等简单问题.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1.实数的大小与实数的运算性质之间的关系设a,b为两个实数,它们在数轴上的点分别记为A,B,如果A落在B的右边,则称a大于b,记为ab;如果A落在B的左边,则称a小于b,记为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】设x=a2b2+5,y=2ab-a2-4a,若xy,则实数a,b应满足的条件为.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目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航2.不等式的基本性质 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航归纳总结(1)对于性质(4)可以看成:若c0,则abacbc;若cbacb,则下列不等式成立的是() 解析:对于选项A,还需有ab0这个前提条件;对于选项B,当a,b都为负数时不成立,或一正一负时可能不成立,如2-3,但22(-3)2不正答案:C 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航【做一做2-2】下列命题中正确的有.若ab,则ac2bc2;答案:目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1.使用不等式的性质时要注意哪些问题?剖析:(1)在应用传递性时,如果两个不等式中有一个带等号,而另一个不带等号,那么等号是不能传递的.如ab,bcabac2bc2;若无c0这个条件,则abac2bc2就错了,因为当c=0时,取等号.(3)ab0anbn0成立的条件是“n为正整数,且n2”.如果去掉这个条件,取n=-1,a=3,b=2,那么就会出现3-12-1,目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航2.比较两数(式)大小的常用方法有哪些?它们有什么区别?剖析:目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四作差比较法 分析:直接作差比较需将展开,过程较为复杂,式子冗长,可以考虑两个式子的特点,根据两个式子的特点,先把式子变形后,再作差比较大小.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四反思当直接作差不容易判断两式的大小或者运算量较大时,可观察式子自身的特点,先变形,再去作差,然后比较大小.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型四题型三不等式的性质【例2】判断下列命题的真假,并简述理由.(1)ab,cda-cb-d;分析:要判断上述命题的真假,依据就是实数的基本性质及实数运算的符号法则,以及不等式的基本性质,经过合理的逻辑推理即可判断.也可令式中字母取一些特殊值,以检验不等式是否成立.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型四题型三解:(1)假命题.理由:令a=5,b=4,c=3,d=1,有ab,cd,但a-cb0|a|nbn,但|a|n与an可能相等,也可能互为相反数,故(4)为假命题,如a=-2,b=1,n=3时,|a|b0,但a3=-8b0,cd0,e0.证明:cd-d0.ab0,a-cb-d0. (*)目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四反思(1)证明不等式的常用方法:直接利用不等式的性质.最常用的性质有传递性、可乘性、同向可加性等;作差法或作商法;函数的单调性.(2)在直接利用不等式的性质证明时,特别注意以下几点:是否是同向不等式;此性质是否可以逆用.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四易错辨析易错点:由于多次应用同向不等式相加(乘)法则导致变量的取值范围扩大.【例4】已知f(x)=mx2-n,且-4f(1)-1,-1f(2)5,求f(3)的取值范围.加减消元,得0m3,1n7,从而,得-7f(3)=9m-n26,即f(3)的取值范围是-7,26.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四错因分析:m,n是两个相互关系、相互制约的量,由条件中的不等式通过加减消元在得出0m3,1n7后,并不意味着m,n可以取得0,3及1,7上的一切值.如当m=0,n=7时,m-n=-7已不满足-4m-n-1.此类题一般是先运用待定系数法把f(3)用f(1),f(2)表示出来,再利用不等式的性质求f(3)的范围.切勿像错解那样先求出m,n的范围,再求f(3)的范围,这样会造成变量的取值范围扩大.目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航题型一题型二题型三题型四目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 51若a0,-1babab2B.ab2abaC.abaab2D.abab2a解析:a0,-1b0,bb2ab2a.答案:D目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 52若ab,则下列不等式中一定成立的是() 答案:C 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航3已知a,b,c均为实数,下面四个命题中正确命题的个数是()A.0B.1C.2D.31 2 3 4 5答案:C 目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 54实数a,b,c,d满足三个条件:dc;a+b=c+d;a+dc,知bdca.答案:bdca目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航ZHISHISHULI知识梳理ZHONGNANJUJIAO重难聚焦SUITANGLIANXI随堂演练DIANLITOUXI典例透析目标导航1 2 3 4 55使不等式a2blg(a-b)0,2a2b+1都成立的a与b的关系是.解析:由条件可知,a与b同时满足|a|b|a-b1,ab+1.故有ab+1,且b0.答案:ab+1,且b0