新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專題17等式與不等式綜合問題（單選+填空）（解析版）

試卷第=page11頁，共=sectionpages33頁專題17等式與不等式綜合問題（單選+填空）一、單選題1．（2023秋·江蘇揚(yáng)州·高三儀征中學(xué)校聯(lián)考期末）已知SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．9 B．10 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由“1”的妙用和基本不等式可求得結(jié)果.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，等號(hào)成立.結(jié)合SKIPIF1<0可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最小值SKIPIF1<0.故選：D.2．（2023春·浙江寧波·高三校聯(lián)考階段練習(xí)）非零實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0成等差數(shù)列，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】B【分析】根據(jù)SKIPIF1<0成等差數(shù)列，可將SKIPIF1<0用SKIPIF1<0表示，再將所求化簡(jiǎn)，利用基本不等式即可得解.【詳解】因?yàn)镾KIPIF1<0成等差數(shù)列，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，取等號(hào)，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：B.3．（2023秋·廣東潮州·高三統(tǒng)考期末）正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，且不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)基本不等式“1”的妙用可得SKIPIF1<0的最小值為4，再根據(jù)含參不等式恒成立解一元二次不等式，即可得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0且SKIPIF1<0時(shí)，等號(hào)成立，則SKIPIF1<0時(shí)，SKIPIF1<0取到最小值4，要使不等式SKIPIF1<0恒成立，即SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.4．（2023春·湖南長沙·高三湖南師大附中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，正數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用SKIPIF1<0可得SKIPIF1<0，由此可化簡(jiǎn)所求式子，結(jié)合基本不等式可求得最小值.【詳解】SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，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：B.5．（2023·湖南岳陽·統(tǒng)考一模）已知正實(shí)數(shù)x，y滿足SKIPIF1<0，則下列不等式恒成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用特殊值判斷AC，利用不等式性質(zhì)及指數(shù)函數(shù)單調(diào)性判斷B，根據(jù)排除法判斷D.【詳解】取SKIPIF1<0，則SKIPIF1<0不成立，故A錯(cuò)誤；由SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故B錯(cuò)誤；取SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，故C錯(cuò)誤；由ABC錯(cuò)誤，排除法知，故D正確.故選：D6．（2023·山東菏澤·統(tǒng)考一模）設(shè)實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，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【分析】分為SKIPIF1<0與SKIPIF1<0，去掉絕對(duì)值后，根據(jù)“1”的代換，化簡(jiǎn)后分別根據(jù)基本不等式，即可求解得出答案.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，此時(shí)有最小值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，此時(shí)有最小值SKIPIF1<0.所以，SKIPIF1<0的最小值為SKIPIF1<0.故選：A.7．（2023·山東濰坊·?？家荒＃┤粽龑?shí)數(shù)a，b滿足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<0【答案】D【分析】根據(jù)函數(shù)單調(diào)性及SKIPIF1<0得到SKIPIF1<0或SKIPIF1<0，分別討論兩種情況下四個(gè)選項(xiàng)是否正確，A選項(xiàng)可以用對(duì)數(shù)函數(shù)單調(diào)性得到，B選項(xiàng)可以用作差法，C選項(xiàng)用作差法及指數(shù)函數(shù)單調(diào)性進(jìn)行求解，D選項(xiàng)，需要構(gòu)造函數(shù)進(jìn)行求解.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0為單調(diào)遞增函數(shù)，故SKIPIF1<0，由于SKIPIF1<0，故SKIPIF1<0，或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0；SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0；SKIPIF1<0，SKIPIF1<0；故ABC均錯(cuò)誤；D選項(xiàng)，SKIPIF1<0，兩邊取自然對(duì)數(shù)，SKIPIF1<0，因?yàn)椴还躍KIPIF1<0，還是SKIPIF1<0，均有SKIPIF1<0，所以SKIPIF1<0，故只需證SKIPIF1<0即可，設(shè)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），則SKIPIF1<0，令SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0且SKIPIF1<0上恒成立，故SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）單調(diào)遞減，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，結(jié)論得證，D正確故選：D8．（2023秋·河北邢臺(tái)·高三統(tǒng)考期末）若SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0的最小值為SKIPIF1<0 B．SKIPIF1<0的最小值為SKIPIF1<0C．SKIPIF1<0的最小值為16 D．SKIPIF1<0沒有最小值【答案】A【分析】先將題意整理成SKIPIF1<0，然后利用基本不等式可得到SKIPIF1<0，最后檢驗(yàn)SKIPIF1<0是否成立即可【詳解】由SKIPIF1<0，得SKIPIF1<0.因?yàn)镾KIPIF1<0，所以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.由SKIPIF1<0得SKIPIF1<0,設(shè)函數(shù)SKIPIF1<0，則由SKIPIF1<0，得SKIPIF1<0在SKIPIF1<0上至少一個(gè)零點(diǎn)，此時(shí)SKIPIF1<0，故存在SKIPIF1<0，使得不等式SKIPIF1<0中的等號(hào)成立，故SKIPIF1<0的最小值為SKIPIF1<0.故選：A【點(diǎn)睛】關(guān)鍵點(diǎn)睛：這道題關(guān)鍵的地方在于檢驗(yàn)SKIPIF1<0是否成立，需要構(gòu)造SKIPIF1<0，并結(jié)合零點(diǎn)存在定理進(jìn)行驗(yàn)證9．（2023秋·河北邯鄲·高三統(tǒng)考期末）已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．10 B．9 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由已知，可設(shè)SKIPIF1<0，SKIPIF1<0，利用換底公式表示出SKIPIF1<0，SKIPIF1<0帶入SKIPIF1<0中，得到m，n的等量關(guān)系，然后利用“1”的代換借助基本不等式即可求解最值.【詳解】由已知，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，代入SKIPIF1<0得：SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)等號(hào)成立.SKIPIF1<0的最小值為SKIPIF1<0.故選：C.10．（2023春·浙江紹興·高三統(tǒng)考開學(xué)考試）對(duì)于任意實(shí)數(shù)SKIPIF1<0及SKIPIF1<0，均有SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先將除了SKIPIF1<0以外的量SKIPIF1<0看成常量，運(yùn)用基本不等式先求出左邊表達(dá)式的最小值，然后利用分離參數(shù)，結(jié)合對(duì)勾函數(shù)性質(zhì)求解.【詳解】由基本不等式，SKIPIF1<0，故只需要SKIPIF1<0即可，即對(duì)于任意的SKIPIF1<0，SKIPIF1<0恒成立，等價(jià)于對(duì)任意的SKIPIF1<0，SKIPIF1<0，或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，原式可變形為SKIPIF1<0，記SKIPIF1<0，根據(jù)對(duì)勾函數(shù)性質(zhì)SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，于是SKIPIF1<0在SKIPIF1<0上遞增，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，原式可變形為SKIPIF1<0，記SKIPIF1<0，根據(jù)對(duì)勾函數(shù)性質(zhì)SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，于是SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，當(dāng)SKIPIF1<0，當(dāng)SKIPIF1<0，注意到SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0.綜上，SKIPIF1<0.故選：D11．（2023·浙江·統(tǒng)考一模）若SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】先把已知整理成SKIPIF1<0的形式，再把等式的右邊利用柯西不等式進(jìn)行放縮，得到關(guān)于SKIPIF1<0的一元二次不等式進(jìn)行求解.【詳解】由已知SKIPIF1<0整理得SKIPIF1<0，由柯西不等式得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：C．二、填空題12．（2023秋·江蘇揚(yáng)州·高三校聯(lián)考期末）已知關(guān)于SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則SKIPIF1<0的最小值為______.【答案】SKIPIF1<0【分析】根據(jù)不等式分類討論分析可知，SKIPIF1<0為SKIPIF1<0的零點(diǎn)，可得方程，運(yùn)算整理結(jié)合基本不等式求值.【詳解】SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0；由SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0為SKIPIF1<0的零點(diǎn)，∴SKIPIF1<0，SKIPIF1<0.∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.故答案為：SKIPIF1<0.13．（2023秋·江蘇揚(yáng)州·高三?？计谀┮阎猄KIPIF1<0，SKIPIF1<0，SKIPIF1<0是正實(shí)數(shù)，且SKIPIF1<0，則SKIPIF1<0最小值為__________.【答案】SKIPIF1<0【分析】由于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是正實(shí)數(shù)，且SKIPIF1<0，所以先結(jié)合基本不等式“1”的代換求SKIPIF1<0的最小值，得SKIPIF1<0，則SKIPIF1<0，再根據(jù)基本不等式湊項(xiàng)法求SKIPIF1<0的最小值，即可求得SKIPIF1<0的最小值.【詳解】解：SKIPIF1<0，由于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是正實(shí)數(shù)，且SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0時(shí)等號(hào)成立，則SKIPIF1<0的最小值為SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，則SKIPIF1<0最小值為SKIPIF1<0.故答案為：SKIPIF1<0.14．（2023春·江蘇揚(yáng)州·高三揚(yáng)州市新華中學(xué)?？奸_學(xué)考試）若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0SKIPIF1<0處的切線也是曲線SKIPIF1<0的切線，則SKIPIF1<0的最小值為_____．【答案】SKIPIF1<0【分析】由兩條曲線的公切線斜率分別等于各曲線上切點(diǎn)處的導(dǎo)數(shù)值，以及各曲線上切點(diǎn)分別滿足切線方程來列方程組，得到SKIPIF1<0與SKIPIF1<0滿足的關(guān)系式，將原式中的SKIPIF1<0替換，再利用基本不等式求最小值即可.【詳解】曲線SKIPIF1<0在點(diǎn)A處的切線可寫作SKIPIF1<0設(shè)該切線在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則有SKIPIF1<0，消去t得SKIPIF1<0則SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取得該最小值.故答案為：SKIPIF1<0.15．（2023春·浙江杭州·高三浙江省杭州第二中學(xué)?？茧A段練習(xí)）已知正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值是___________.【答案】SKIPIF1<0【分析】根據(jù)不等式特征可通過構(gòu)造函數(shù)SKIPIF1<0，利用函數(shù)單調(diào)性解不等式可得SKIPIF1<0，再根據(jù)基本不等式即可求得SKIPIF1<0的最小值是SKIPIF1<0.【詳解】由題意可得將不等式變形成SKIPIF1<0；又因?yàn)镾KIPIF1<0都是正數(shù)，所以SKIPIF1<0；可構(gòu)造函數(shù)SKIPIF1<0，易知函數(shù)為增函數(shù)，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，根據(jù)函數(shù)單調(diào)性可得SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0取等號(hào)，因此SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<016．（2023·浙江·永嘉中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知實(shí)數(shù)SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的最小值是______.【答案】9【分析】將已知條件SKIPIF1<0通過恒等變形，再利用基本不等式即可求解.【詳解】由已知條件得SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.故答案為：9.17．（2023春·廣東江門·高三校聯(lián)考開學(xué)考試）已知正數(shù)x，y，z滿足SKIPIF1<0，當(dāng)SKIPIF1<0取最大值時(shí)，SKIPIF1<0的最小值為______.【答案】SKIPIF1<0##SKIPIF1<0【分析】由條件化簡(jiǎn)SKIPIF1<0，結(jié)合基本不等式求其最大值，確定取最大值的條件，再結(jié)合二次函數(shù)性質(zhì)求SKIPIF1<0的最小值.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，所以當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0取最大值SKIPIF1<0，所以當(dāng)SKIPIF1<0取最大值時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0.故答案為：SKIPIF1<0.18．（2023秋·廣東·高三校聯(lián)考期末）已知a，b都是正數(shù)，則SKIPIF1<0的最小值是______．【答案】2【分析】設(shè)SKIPIF1<0，SKIPIF1<0，解出SKIPIF1<0，SKIPIF1<0，代入化簡(jiǎn)得SKIPIF1<0，利用基本不等式即可求出最值.【詳解】因?yàn)镾KIPIF1<0均為正實(shí)數(shù)，故設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0聯(lián)立解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，故答案為：2.19．（2023秋·湖南長沙·高三校考階段練習(xí)）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，若存在兩項(xiàng)SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的最小值為_____________.【答案】SKIPIF1<0【分析】先根據(jù)SKIPIF1<0可得數(shù)列SKIPIF1<0是首項(xiàng)為SKIPIF1<0，公比為SKIPIF1<0的等比數(shù)列，即可得到SKIPIF1<0，結(jié)合SKIPIF1<0可得SKIPIF1<0，再結(jié)合基本不等式求解即可.【詳解】由SKIPIF1<0，得SKIPIF1<0，兩式相減得SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，所以數(shù)列SKIPIF1<0是首項(xiàng)為SKIPIF1<0，公比為SKIPIF1<0的等比數(shù)列，即SKIPIF1<0.又SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào).所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.20．（2023春·湖北荊州·高三沙市中學(xué)?？茧A段練習(xí)）若SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0的最小值為m，SKIPIF1<0的最大值為n，則mn為___________，【答案】SKIPIF1<0【分析】根據(jù)條件等式利用基本不等式中“1”的妙用可求得SKIPIF1<0，由SKIPIF1<0并結(jié)合SKIPIF1<0即可求得SKIPIF1<0，便可得出SKIPIF1<0.【詳解】由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立；即SKIPIF1<0的最小值為SKIPIF1<0；SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立；即SKIPIF1<0的最大值為SKIPIF1<0；所以SKIPIF1<0.故答案為：SKIPIF1<021．（2023春·湖北武漢·高三華中師大一附中?？茧A段練習(xí)）已知正數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【分析】利用基本不等式知SKIPIF1<0，令SKIPIF1<0，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性可知SKIPIF1<0，進(jìn)而可得SKIPIF1<0，結(jié)合已知可得SKIPIF1<0，由取等條件即可求解.【詳解】因?yàn)閍，b都為正數(shù)，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，求導(dǎo)SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；可知SKIPIF1<0在SKIPIF1<0處取得最大值，故SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查利用基本不等式及利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性證明不等式，解題的關(guān)鍵是構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，從而證得SKIPIF1<0，再結(jié)合基本不等式及取等條件即可求解，考查學(xué)生的轉(zhuǎn)化能力與運(yùn)算求解能力，屬于難題.22．（2023·山東濟(jì)寧·統(tǒng)考一模）已知函數(shù)SKIPIF1<0且SKIPIF1<0的圖象過定點(diǎn)A，且點(diǎn)A在直線SKIPIF1<0上，則SKIPIF1<0的最小值是______.【答案】SKIPIF1<0【分析】求出函數(shù)所過的定點(diǎn)SKIPIF1<0，則有SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，化簡(jiǎn)整理，分離常數(shù)再結(jié)合基本不等式求解即可.【詳解】函數(shù)SKIPIF1<0且SKIPIF1<0的圖象過定點(diǎn)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0時(shí)，取等號(hào)，所以SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<0.23．（2023秋·河北·高三統(tǒng)考階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的最小值為___________．【答案】SKIPIF1<0【分析】根據(jù)已知條件變形，再應(yīng)用基本不等式求最小值即可.【詳解】SKIPIF1<0則SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立．SKIPIF1<0，∴SKIPIF1<0最小為SKIPIF1<0，此時(shí)SKIPIF1<0．故答案為:SKIPIF1<0.24．（2023·重慶渝中·高三重慶巴蜀中學(xué)?？茧A段練習(xí)）已知實(shí)數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0的最大值為_____________．【答案】2【分析】先消元，再用基本不等式即可求出最大值.【詳解】由SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，SKIPIF1<0，或者SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0的最大值為2．故答案為：2.25．（2023·重慶·統(tǒng)考一模）已知SKIPIF1<0，則SKIPIF1<0的最小值是___________.【答案】4【分析】把SKIPIF1<0化為SKIPIF1<0，再利用“1”的妙用，結(jié)合基本不等式即可得到答案.【詳解】SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，取等號(hào)，故SKIPIF1<0的最小值是4，故答案為：SKIPIF1<0.26．（2023·黑龍江哈爾濱·哈爾濱三中?？家荒＃┮阎猄KIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為______．【答案】2【分析】根據(jù)基本不等式湊項(xiàng)法和“1”的巧用即可求得最值.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.27．（2023秋·山西長治·高三校聯(lián)考階段練習(xí)）已知兩個(gè)正數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為__________.【答案】2【分析】先通分得到等式SKIPIF1<0，再通過配方法得到等式SKIPIF1<0，最后通過基本不等式得到SKIPIF1<0的取值范圍即可得到答案.【詳解】由SKIPIF1<0得SKIPIF1<0.所以SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0的最大值為2.故答案為：2.28．（2023·吉林·長春十一高校聯(lián)考模擬預(yù)測(cè)）已知正實(shí)數(shù)x，y滿足SKIPIF1<0，則SKIPIF1<0的小值為______．【答案】SKIPIF1<0【分析】利用待定系數(shù)法可得出SKIPIF1<0，與SKIPIF1<0相乘，展開后利用基本不等式可求得SKIPIF1<0的最小值.【詳解】設(shè)SKIPIF1<0，可得SKIPIF1<