新高考數(shù)學(xué)一輪復(fù)習(xí)提升訓(xùn)練2.2 基本不等式（精講）（解析版）

2.2基本不等式（精講）（提升版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一基本不等式?？夹问健纠?-1】（2022·河北石家莊·高三階段練習(xí)）（多選）已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0的最小值是1 B．SKIPIF1<0的最小值是SKIPIF1<0C．SKIPIF1<0的最小值是4 D．SKIPIF1<0的最小值是5【答案】BC【解析】由已知，得SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最大值是SKIPIF1<0，所以選項(xiàng)A錯(cuò)誤；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值是SKIPIF1<0，所以選項(xiàng)B正確；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值是4，所以選項(xiàng)C正確；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值是SKIPIF1<0，所以選項(xiàng)D錯(cuò)誤．故選：BC．【例1-2】（2022·全國(guó)·模擬預(yù)測(cè)）已知a，b為非負(fù)數(shù)，且滿足SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．40 B．SKIPIF1<0 C．42 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0.故選：D【例1-3】（2022·全國(guó)·高三專題練習(xí)）已知正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．9 B．SKIPIF1<0 C．10 D．無(wú)最小值【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，即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的最小值為SKIPIF1<0，故選：A【例1-4】（2022·全國(guó)·高三專題練習(xí)）已知實(shí)數(shù)SKIPIF1<0，滿足SKIPIF1<0，若不等式SKIPIF1<0對(duì)任意的正實(shí)數(shù)SKIPIF1<0恒成立，那么實(shí)數(shù)m的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】D【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，∵

SKIPIF1<0∴

SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，∵不等式SKIPIF1<0對(duì)任意的正實(shí)數(shù)SKIPIF1<0恒成立，∴SKIPIF1<0，故選：D.【一隅三反】1．（2022·海南）（多選）已知SKIPIF1<0，SKIPIF1<0是正實(shí)數(shù)，則下列選項(xiàng)正確的是（

）A．若SKIPIF1<0，則SKIPIF1<0有最小值2B．若SKIPIF1<0，則SKIPIF1<0有最大值5C．若SKIPIF1<0，則SKIPIF1<0有最大值SKIPIF1<0D．SKIPIF1<0有最小值SKIPIF1<0【答案】AC【解析】對(duì)于A，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，則SKIPIF1<0有最小值2，故A正確；對(duì)于B，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，則SKIPIF1<0有最大值4，故B錯(cuò)誤；對(duì)于C，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，則則SKIPIF1<0有最大值SKIPIF1<0，故C正確；對(duì)于D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故D錯(cuò)誤；故選：AC2．（2022·全國(guó)·高三專題練習(xí)（理））若a，b，c均為正實(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【解析】因?yàn)閍，b均為正實(shí)數(shù)，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，則SKIPIF1<0的最大值為SKIPIF1<0．故選：A．3．（2022·全國(guó)·高三專題練習(xí)）已知三次函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0恒成立，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)），SKIPIF1<0，即SKIPIF1<0的最小值為SKIPIF1<0.故選：SKIPIF1<0.4．（2022·全國(guó)·高三專題練習(xí)）若兩個(gè)正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0且存在這樣的SKIPIF1<0，SKIPIF1<0使不等式SKIPIF1<0有解，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】C【解析】由SKIPIF1<0知，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，則使不等式SKIPIF1<0有解，只需滿足SKIPIF1<0即可，解得SKIPIF1<0故選：C5．（2022·全國(guó)·高三專題練習(xí)）若實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)_______.【答案】SKIPIF1<0【解析】由SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，其中SKIPIF1<0.則SKIPIF1<0，從而SKIPIF1<0，記SKIPIF1<0，則SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，即最大值為SKIPIF1<0.故答案為：SKIPIF1<0.考點(diǎn)二基本不等式與其他知識(shí)綜合【例2-1】（2022·河南許昌）若直線SKIPIF1<0被圓SKIPIF1<0截得的弦長(zhǎng)為SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】圓的標(biāo)準(zhǔn)方程為SKIPIF1<0，圓心為SKIPIF1<0，半徑為SKIPIF1<0，若直線被截得弦長(zhǎng)為SKIPIF1<0，說(shuō)明圓心在直線：SKIPIF1<0上，即SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立．故選：D.【例2-2】．（2022·全國(guó)·高三專題練習(xí)）設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0的最大值為_(kāi)__________．【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0等號(hào)成立．故最大值為SKIPIF1<0．【例2-3】（2022·山東·廣饒一中）直角三角形SKIPIF1<0中，SKIPIF1<0是斜邊SKIPIF1<0上一點(diǎn)，且滿足SKIPIF1<0，點(diǎn)SKIPIF1<0、SKIPIF1<0在過(guò)點(diǎn)SKIPIF1<0的直線上，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列結(jié)論錯(cuò)誤的是（

）A．SKIPIF1<0為常數(shù) B．SKIPIF1<0的最小值為SKIPIF1<0C．SKIPIF1<0的最小值為SKIPIF1<0 D．SKIPIF1<0、SKIPIF1<0的值可以為SKIPIF1<0，SKIPIF1<0【答案】B【解析】如下圖所示：由SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0、SKIPIF1<0、SKIPIF1<0三點(diǎn)共線，SKIPIF1<0，SKIPIF1<0，故A正確；所以SKIPIF1<0，SKIPIF1<0時(shí)，也滿足SKIPIF1<0，則D選項(xiàng)正確；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，C成立；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，故B選項(xiàng)錯(cuò)誤.故選：B【一隅三反】1．（2022·江西·臨川一中）已知SKIPIF1<0是正實(shí)數(shù)，函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．9 C．SKIPIF1<0 D．2【答案】B【解析】由函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0．SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取到等號(hào)．故選：B．2．（2022·江西·模擬預(yù)測(cè)（理））在正項(xiàng)等比數(shù)列SKIPIF1<0中，SKIPIF1<0，前三項(xiàng)的和為7，若存在m，SKIPIF1<0使得SKIPIF1<0，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)等比數(shù)列SKIPIF1<0的公比為q，SKIPIF1<0前三項(xiàng)的和為7，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），又由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，且m，SKIPIF1<0，故選：B3．（2022·安徽省舒城中學(xué)）如圖，在SKIPIF1<0中，SKIPIF1<0是線段SKIPIF1<0上的一點(diǎn)，且SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線分別交直線SKIPIF1<0，SKIPIF1<0于點(diǎn)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由條件可得SKIPIF1<0，∵SKIPIF1<0∴SKIPIF1<0，因?yàn)镾KIPIF1<0三點(diǎn)共線，∴SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，則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SKIPIF1<0；故選：C．4．（2022·廣東·廣州六中高一期末）己知第二象限角SKIPIF1<0的終邊上有異于原點(diǎn)的兩點(diǎn)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．3 C．SKIPIF1<0 D．4【答案】B【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又第二象限角SKIPIF1<0的終邊上有異于原點(diǎn)的兩點(diǎn)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故選：B5．（2021·江蘇·揚(yáng)州大學(xué)附屬中學(xué)）不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)___________.【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，即不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，要使得SKIPIF1<0有意義，此時(shí)SKIPIF1<0，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，若不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.綜上所述，SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0.6．（2022·安徽·合肥一中）已知圓SKIPIF1<0的半徑為3，SKIPIF1<0，SKIPIF1<0為該圓的兩條切線，SKIPIF1<0為切點(diǎn)，則SKIPIF1<0的最小值為_(kāi)__________.【答案】SKIPIF1<0【解析】如圖所示，設(shè)SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)等號(hào)成立，∴SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<0．7．（2021·四川達(dá)州·一模（文））定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.設(shè)SKIPIF1<0在SKIPIF1<0上最小值為SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，取等號(hào)；所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；又SKIPIF1<0所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，所以SKIPIF1<0；又SKIPIF1<0在SKIPIF1<0上最小值為SKIPIF1<0，所以SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0所以SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng)，3為公差的等差數(shù)列，即SKIPIF1<0所以SKIPIF1<0.故答案為：SKIPIF1<0.考點(diǎn)三連用兩次基本不等式【例3】（2021·廣東河源·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0