新高考數(shù)學(xué)二輪復(fù)習(xí)講義專(zhuān)題03 不等式（原卷版）

專(zhuān)題03講：不等式【考點(diǎn)專(zhuān)題】1．兩個(gè)實(shí)數(shù)比較大小的方法(1)作差法eq\b\lc\{\rc\(\a\vs4\al\co1(a－b>0?a>b,a－b＝0?a＝b,a－b<0?a<b))(a，b∈R)(2)作商法eq\b\lc\{\rc\(\a\vs4\al\co1(\f(a,b)>1?a>b,\f(a,b)＝1?a＝b,\f(a,b)<1?a<b))(a∈R，b>0)2．不等式的基本性質(zhì)性質(zhì)性質(zhì)內(nèi)容特別提醒對(duì)稱(chēng)性a>b?b<a?傳遞性a>b，b>c?a>c?可加性a>b?a＋c>b＋c?可乘性eq\b\lc\\rc\}(\a\vs4\al\co1(a>b,c>0))?ac>bc注意c的符號(hào)eq\b\lc\\rc\}(\a\vs4\al\co1(a>b,c<0))?ac<bc同向可加性eq\b\lc\\rc\}(\a\vs4\al\co1(a>b,c>d))?a＋c>b＋d?同向同正可乘性eq\b\lc\\rc\}(\a\vs4\al\co1(a>b>0,c>d>0))?ac>bd?可乘方性a>b>0?an>bn(n∈N，n≥1)a，b同為正數(shù)可開(kāi)方性a>b>0?eq\r(n,a)>eq\r(n,b)(n∈N，n≥2)a，b同為正數(shù)3．一元二次不等式的解集判別式Δ＝b2－4acΔ>0Δ＝0Δ<0二次函數(shù)y＝ax2＋bx＋c(a>0)的圖象方程ax2＋bx＋c＝0(a>0)的根有兩相異實(shí)根x1，x2(x1<x2)有兩相等實(shí)根x1＝x2＝－eq\f(b,2a)沒(méi)有實(shí)數(shù)根ax2＋bx＋c>0(a>0)的解集{x|x<x1或x>x2}eq\b\lc\{\rc\}(\a\vs4\al\co1(x\b\lc\|\rc\(\a\vs4\al\co1(x≠－\f(b,2a))))){x|x∈R}ax2＋bx＋c<0(a>0)的解集{x|x1<x<x2}??4．基本不等式eq\r(ab)≤eq\f(a＋b,2)(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eq\f(a＋b,2)叫做正數(shù)a，b的算術(shù)平均數(shù)，eq\r(ab)叫做正數(shù)a，b的幾何平均數(shù)．5．幾個(gè)重要的不等式(1)a2＋b2≥2ab(a，b∈R)．(2)eq\f(b,a)＋eq\f(a,b)≥2(a，b同號(hào))．(3)ab≤eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))2(a，b∈R)．(4)eq\f(a2＋b2,2)≥eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))2(a，b∈R)．以上不等式等號(hào)成立的條件均為a＝b.6．用基本不等式求最值用基本不等式eq\r(ab)≤eq\f(a＋b,2)求最值應(yīng)注意：一正二定三相等.(1)a，b是正數(shù)；(2)①如果ab等于定值P，那么當(dāng)a＝b時(shí)，和a＋b有最小值2eq\r(P)；②如果a＋b等于定值S，那么當(dāng)a＝b時(shí)，積ab有最大值eq\f(1,4)S2.(3)討論等號(hào)成立的條件是否滿(mǎn)足．【方法技巧】一、比較大小的常用方法(1)作差法：①作差；②變形；③定號(hào)；④得出結(jié)論．(2)作商法：①作商；②變形；③判斷商與1的大小關(guān)系；④得出結(jié)論．(3)構(gòu)造函數(shù)，利用函數(shù)的單調(diào)性比較大?。?、判斷不等式的常用方法(1)直接利用不等式的性質(zhì)逐個(gè)驗(yàn)證，利用不等式的性質(zhì)判斷不等式是否成立時(shí)要特別注意前提條件．(2)利用特殊值法排除錯(cuò)誤答案．(3)利用函數(shù)的單調(diào)性，當(dāng)直接利用不等式的性質(zhì)不能比較大小時(shí)，可以利用指數(shù)函數(shù)、對(duì)數(shù)函數(shù)、冪函數(shù)等函數(shù)的單調(diào)性來(lái)比較．三、利用基本不等式求最值(1)前提：“一正”“二定”“三相等”．(2)要根據(jù)式子的特征靈活變形，配湊出積、和為常數(shù)的形式，然后再利用基本不等式．(3)條件最值的求解通常有三種方法：一是配湊法；二是將條件靈活變形，利用常數(shù)“1”代換的方法；三是消元法．【核心題型】題型一：比較兩個(gè)數(shù)（式）的大小1．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．已知：SKIPIF1<0，則3，SKIPIF1<0，SKIPIF1<0的大小關(guān)系是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．無(wú)法確定題型二：不等式的基本性質(zhì)4．對(duì)于實(shí)數(shù)a，b，c，下列命題中正確的是（

）A．若SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<05．（多選）已知a，b，c，d均為實(shí)數(shù)，則下列命題正確的是（

）A．若a>b，c>d，則a-d>b-c B．若a>b，c>d則ac>bdC．若ab>0，bc-ad>0，則SKIPIF1<0 D．若a>b，c>d>0，則SKIPIF1<06．（多選）已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型三：不等式性質(zhì)的綜合應(yīng)用7．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．已知SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的取值范圍為_(kāi)_________.題型四：利用基本不等式求最值命題點(diǎn)1配湊法10．設(shè)實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．611．已知x>0，y>0,2x＋3y＝6，則xy的最大值為_(kāi)_______．12．已知a>b>c，求(a－c)eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,a－b)＋\f(1,b－c)))的最小值．命題點(diǎn)2常數(shù)代換法13．已知SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．7 B．SKIPIF1<0 C．4 D．SKIPIF1<014．已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．9 B．10 C．11 D．SKIPIF1<015．若實(shí)數(shù)SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．216．已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)________．命題點(diǎn)3消元法17．負(fù)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<018．若實(shí)數(shù)x，y滿(mǎn)足xy＋3x＝3eq\b\lc\(\rc\)(\a\vs4\al\co1(0<x<\f(1,2)))，則eq\f(3,x)＋eq\f(1,y－3)的最小值為_(kāi)_______．19．已知SKIPIF1<0SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2題型五：基本不等式的綜合應(yīng)用20．已知正實(shí)數(shù)a、b滿(mǎn)足SKIPIF1<0，若SKIPIF1<0的最小值為4，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<021．在SKIPIF1<0中，角SKIPIF1<0所對(duì)的邊分別為SKIPIF1<0，且點(diǎn)SKIPIF1<0滿(mǎn)足SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<022．設(shè)等差數(shù)列{an}的公差為d，其前n項(xiàng)和是Sn，若a1＝d＝1，則eq\f(Sn＋8,an)的最小值是________．【高考必刷】一、單選題1．（2021·山西太原·高一階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0和SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<02．（2022·湖北·葛洲壩中學(xué)高一階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．不能確定3．（2022·江蘇宿遷·高一期中）若SKIPIF1<0且SKIPIF1<0，則下列不等式一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·江西·貴溪市實(shí)驗(yàn)中學(xué)高三階段練習(xí)（文））若SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·黑龍江·牡丹江市第三高級(jí)中學(xué)高三階段練習(xí)）已知SKIPIF1<0為正實(shí)數(shù)且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．36．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知兩個(gè)正實(shí)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．8 D．37．（2022·全國(guó)·高一單元測(cè)試）已知正數(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<08．（2022·浙江·高一期中）已知實(shí)數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．6 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2021·安徽合肥·高一期末）已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2022·黑龍江·哈爾濱市第一中學(xué)校高一階段練習(xí)）下列說(shuō)法正確的是（

）A．若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0至少有一個(gè)大于2B．SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<011．（2022·甘肅省會(huì)寧縣第一中學(xué)高一期中）若函數(shù)SKIPIF1<0在SKIPIF1<0處取最小值，則SKIPIF1<0等于（

）A．3 B．SKIPIF1<0 C．SKIPIF1<0 D．412．（2015·湖南·高考真題（文））若實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．413．（2022·山東·青島二中高一期中）十六世紀(jì)中葉，英國(guó)數(shù)學(xué)家雷科德在《礪智石》一書(shū)中首先把“=”作為等號(hào)使用，后來(lái)英國(guó)資學(xué)家哈利奧特首次使用“>”和“<”符號(hào)，并逐步被數(shù)學(xué)界接受志不等號(hào)的引入對(duì)不等式的發(fā)展景響深遠(yuǎn)．已知a，b為非零實(shí)數(shù)，且SKIPIF1<0；則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<014．（2022·福建·福州第十五中學(xué)高三階段練習(xí)）已知SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<015．（2021·山西·太原市第五十六中學(xué)校高一階段練習(xí)）若正數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，當(dāng)SKIPIF1<0取得最小值時(shí)，SKIPIF1<0的值為（）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．516．（2022·全國(guó)·高三專(zhuān)題練習(xí)）當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<017．（2022·天津·靜海一中高一期中）已知正數(shù)SKIPIF1<0、SKIPIF1<0滿(mǎn)足SKIPIF1<0，若不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2022·福建·莆田一中高一階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的最大值為（

）A．10 B．9 C．8 D．7二、多選題19．（2022·全國(guó)·高一單元測(cè)試）下列命題為真命題的是（

）A．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0 B．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0 D．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<020．（2022·河南省浚縣第一中學(xué)高一階段練習(xí)）若正實(shí)數(shù)a，b滿(mǎn)足SKIPIF1<0則下列說(shuō)法正確的是（

）A．a(chǎn)b有最大值SKIPIF1<0 B．SKIPIF1<0有最大值SKIPIF1<0C．SKIPIF1<0有最小值2 D．SKIPIF1<0有最大值SKIPIF1<0三、填空題21．（2022·山東·乳山市銀灘高級(jí)中學(xué)高一階段練習(xí)）若實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍為_(kāi)_______．22．（2018·天津·高考真題（理））已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)____________.23．（2023·廣東·惠來(lái)縣第一中學(xué)高一期中）已知SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)_______．24．（2022·天津市第四中學(xué)高三期中）已知SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.25．（2019·天津·高考真題（文））設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________.26．（2017·天津·高考真題（文））若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________.27．（2017·江蘇·高考真題）某公司一年購(gòu)買(mǎi)某種貨物SKIPIF1<0噸，每次購(gòu)買(mǎi)SKIPIF1<0噸，運(yùn)費(fèi)為SKIPIF1<0萬(wàn)元/次，一年的總存儲(chǔ)費(fèi)用為SKIPIF1<0萬(wàn)元．要使一年的總運(yùn)費(fèi)與總存儲(chǔ)費(fèi)用之和最小，則SKIPIF1<0的值是__________．28．（2022·全國(guó)·高考真題（理））已知SKIPIF1<0中，點(diǎn)D在邊BC上，SKIPIF1<0．當(dāng)SKIPIF1<0取得最小值時(shí)，SKIPIF1<0________．四、解答題29．（2022·海南·儋州川綿中學(xué)高一期中）比較下列兩組代數(shù)式的大小(1)SKIPIF1<0和SKIPIF1<0(2)SKIPIF1<0與SKIPIF1<030．（2022·河北·衡水市冀州區(qū)滏運(yùn)中學(xué)高一階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0，SKIPIF1