新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第02講 等式性質(zhì)與不等式（講）（解析版）

第2講等式性質(zhì)與不等式本講為高考重要知識(shí)點(diǎn)，題型主要和其他知識(shí)結(jié)合考察，屬于工具型知識(shí)點(diǎn)，梳理等式性質(zhì)的基礎(chǔ)上，通過(guò)類(lèi)比，研究不等式的性質(zhì)，并利用這些性質(zhì)研究一類(lèi)重要的不等式-基本不等式。體會(huì)函數(shù)觀點(diǎn)統(tǒng)一方程和不等式的數(shù)學(xué)思想?？键c(diǎn)一等式性質(zhì)與不等式的性質(zhì)1．實(shí)數(shù)的大小順序與運(yùn)算性質(zhì)的關(guān)系(1)a>b?a－b>0；(2)a＝b?a－b＝0；(3)a<b?a－b<0.2.不等式的性質(zhì)(1)對(duì)稱(chēng)性：a＞b?b＜a；(2)傳遞性：a＞b，b＞c?a＞c；(3)可加性：a＞b?a＋c＞b＋c；a＞b，c＞d?a＋c＞b＋d；(4)可乘性：a＞b，c＞0?ac＞bc；a>b，c<0?ac<bc；a＞b＞0，c＞d＞0?ac＞bd；(5)可乘方：a＞b＞0?an＞bn(n∈N，n≥1)；(6)可開(kāi)方：a＞b＞0?SKIPIF1<0＞SKIPIF1<0(n∈N，n≥2).考點(diǎn)二基本不等式1.基本不等式：SKIPIF1<0≤SKIPIF1<0(1)基本不等式成立的條件：a≥0，b≥0.(2)等號(hào)成立的條件：當(dāng)且僅當(dāng)a＝b時(shí)取等號(hào).(3)其中SKIPIF1<0稱(chēng)為正數(shù)a，b的算術(shù)平均數(shù)，SKIPIF1<0稱(chēng)為正數(shù)a，b的幾何平均數(shù).2.兩個(gè)重要的不等式(1)a2＋b2≥2ab(a，b∈R)，當(dāng)且僅當(dāng)a＝b時(shí)取等號(hào).(2)ab≤SKIPIF1<0(a，b∈R)，當(dāng)且僅當(dāng)a＝b時(shí)取等號(hào).3.利用基本不等式求最值已知x≥0，y≥0，則(1)如果積xy是定值p，那么當(dāng)且僅當(dāng)x＝y(tǒng)時(shí)，x＋y有最小值是2SKIPIF1<0(簡(jiǎn)記：積定和最小).(2)如果和x＋y是定值s，那么當(dāng)且僅當(dāng)x＝y(tǒng)時(shí)，xy有最大值是SKIPIF1<0(簡(jiǎn)記：和定積最大).注意：1.SKIPIF1<0≥2(a，b同號(hào))，當(dāng)且僅當(dāng)a＝b時(shí)取等號(hào).2.ab≤SKIPIF1<0≤SKIPIF1<0.3.SKIPIF1<0(a>0，b>0).高頻考點(diǎn)一等式性質(zhì)與不等式性質(zhì)例1、已知，則下列結(jié)論正確的是

A. B. C. D.【答案】【解析】【解答】解：對(duì)于：當(dāng)時(shí)，根式無(wú)意義，選項(xiàng)錯(cuò)誤；

對(duì)于：在一個(gè)不等式兩邊同時(shí)加上一個(gè)實(shí)數(shù)，不等式仍成立，故B正確；

對(duì)于：，當(dāng)時(shí)，不成立；

對(duì)于：當(dāng)，時(shí)，，但不成立．

故選：．【變式訓(xùn)練】1．若SKIPIF1<0，則下列不等式正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對(duì)于A，若SKIPIF1<0，則SKIPIF1<0，所以A錯(cuò)誤，對(duì)于B，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以B正確，對(duì)于C，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以C錯(cuò)誤，對(duì)于D，若SKIPIF1<0，則SKIPIF1<0，所以D錯(cuò)誤，故選：B高頻考點(diǎn)二“1”的代換型例2、已知x，y均為正實(shí)數(shù)，且SKIPIF1<0，則x+3y的最小值為_(kāi)_________【詳解】x，y均為正實(shí)數(shù),SKIPIF1<0,SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)等號(hào)成立.故答案為：2.【變式訓(xùn)練】1.已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（）A．20 B．24 C．25 D．28【答案】C【詳解】由題意SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立．故選：C．2.已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．13 B．19 C．21 D．27【答案】D【詳解】SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，b＝6時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為27。故選：D3.已知正實(shí)數(shù)SKIPIF1<0，b滿(mǎn)足SKIPIF1<0＋b＝1，則SKIPIF1<0的最小值為_(kāi)____【詳解】因?yàn)镾KIPIF1<0，且SKIPIF1<0都是正實(shí)數(shù).所以SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.所以SKIPIF1<0的最小值為SKIPIF1<0【做題技巧】1.基本公式SKIPIF1<02.一正二定三相等。是均值成立的前提條件。高頻考點(diǎn)三“和”與“積”互消型例3、已知x、y都是正數(shù)，且滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)________．【答案】18.【詳解】因?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取等號(hào)）即SKIPIF1<0，解得SKIPIF1<0，所以得SKIPIF1<0，所以SKIPIF1<0的最大值是SKIPIF1<0.此時(shí)SKIPIF1<0，SKIPIF1<0.故答案為：18.【變式訓(xùn)練】1.已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題可知SKIPIF1<0，乘“SKIPIF1<0”得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取等號(hào)，則SKIPIF1<0的最小值為SKIPIF1<0.故選：A2.已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________．【答案】6【詳解】由SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)．故答案為：6．【基本規(guī)律】1.有“和”、“積”無(wú)常數(shù)，可以同除，化回到“1”的代換型。如變式1；2.有“和”、“積”有常數(shù)求積型，可以借助基本不等式構(gòu)造不等式求解，如典例分析；3..有“和”、“積”有常數(shù)求和型，可以借助基本不等式構(gòu)造不等式求解，如變式2。高頻考點(diǎn)四以分母為主元構(gòu)造型例4、已知非負(fù)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值是（）A．3 B．4 C．10 D．16【答案】B【詳解】由SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0取等號(hào)，故選：B【變式訓(xùn)練】1.已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（）A．9 B．10 C．11 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0的最小值為9．故選：A．2.已知正數(shù)SKIPIF1<0、SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】已知正數(shù)SKIPIF1<0、SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值是SKIPIF1<0.故選：C.3.設(shè)SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．4 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)故選：A【基本規(guī)律】構(gòu)造分母型：1.以分母為主元構(gòu)造，可以直接分母換元，變化后為“1”的代換，如典例分析2.構(gòu)造過(guò)程中，分子會(huì)有分母參數(shù)的變化，可以分離常數(shù)后再構(gòu)造分母，如變式23.變式3是三項(xiàng)構(gòu)造，且無(wú)條件等式。高頻考點(diǎn)五構(gòu)造分母：待定系數(shù)例5、已知正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由正實(shí)數(shù)x,y滿(mǎn)足4x+3y=4,可得2(2x+1)+(3y+2)=8.令a=2x+1,b=3y+2,可得2a+b=8.所求SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以答案為SKIPIF1<0.故選:A.【變式訓(xùn)練】1.知正實(shí)數(shù)SKIPIF1<0、SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，所以，SKIPIF1<0SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：A.2.已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0取到最小值為．【答案】SKIPIF1<0．【解析】試題分析：令SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，即SKIPIF1<0的最小值是SKIPIF1<0．【基本規(guī)律】特征：條件等式和所求式子之間變量系數(shù)“不一致”方法：直觀湊配或者分母換元高頻考點(diǎn)六分子含參型：分離分子型例6、若SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0，

當(dāng)且僅當(dāng)SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1.已知正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因此SKIPIF1<0，因?yàn)镾KIPIF1<0是正實(shí)數(shù)，所以SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，即SKIPIF1<0時(shí)取等號(hào)，即SKIPIF1<0時(shí)取等號(hào)），故選：A2.若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)________【答案】SKIPIF1<0【分析】令SKIPIF1<0，可得SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0，再結(jié)合基本不等式可求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.3.若正實(shí)數(shù)x，y滿(mǎn)足2x+y=2，則SKIPIF1<0的最小值是_____．【答案】SKIPIF1<0【方法總結(jié)】1.分離分子原理題，如典例分析2.分子二次型換元分離，如變式23.分子二次型湊配構(gòu)造分離，如變式3高頻考點(diǎn)七反解代入型:消元法例7、已知正數(shù)SKIPIF1<0，SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)_____．【答案】SKIPIF1<0【詳解】由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，、所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1.已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，記SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0等號(hào)成立，此時(shí)SKIPIF1<0，SKIPIF1<0.2.若正數(shù)SKIPIF1<0，SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值是______，此時(shí)SKIPIF1<0______.【答案】22解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，因?yàn)镾KIPIF1<0、SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，故答案為：2；2．3.若正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.【答案】SKIPIF1<0【詳解】由SKIPIF1<0且SKIPIF1<0知：SKIPIF1<0，∴SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，即SKIPIF1<0時(shí)等號(hào)成立.故答案為：SKIPIF1<0【方法總結(jié)】條件等式和所求等式之間互化難以實(shí)現(xiàn)，可以借助反解代入消元，再重新構(gòu)造。高頻考點(diǎn)八反解代入型:消元法例8、非負(fù)實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.【答案】SKIPIF1<0【詳解】由題意，非負(fù)實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，可得SKIPIF1<0,又由SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.【變式訓(xùn)練】1.已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值是___．【答案】SKIPIF1<0【解析】原式可變形為SKIPIF1<0，兩邊同時(shí)乘以2，得SKIPIF1<0，所以SKIPIF1<0，即x+2ySKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立。填SKIPIF1<02.已知a,b∈R+，且(a+b)(a+2b)+a+b=9，則【答案】6【詳解】a,b∈R+，且(a+b)(a+2b)+a+b=9，即有（a+b）（a+2b+1）=9，

即（2a+2b）（a+2b+1）=18，可得3a+4b+1=（2a+2b）+（a+2b+1）≥22a+2ba+2b+1=62，

當(dāng)且僅當(dāng)2a+2b=a+2b+1【方法總結(jié)】特征：條件式子復(fù)雜，一般有一次和二次（因式分解展開(kāi)就是一次和二次），可能就符合因式分解原理高頻考點(diǎn)九均值用兩次例9、SKIPIF1<0是不同時(shí)為0的實(shí)數(shù)，則SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)閍，b均為正實(shí)數(shù)，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<