人教A版高中數(shù)學(xué)(必修第一冊(cè))同步講義第14講 拓展二 基本不等式與對(duì)勾函數(shù)（含解析）

第07講拓展二基本不等式與對(duì)勾函數(shù)一、知識(shí)清單1、基本不等式常用技巧利用基本不等式求最值的變形技巧——湊、拆（分子次數(shù)高于分母次數(shù)）、除（分子次數(shù)低于分母次數(shù)）、代（1的代入）、解（整體解）.①湊：湊項(xiàng)，例：SKIPIF1<0；湊系數(shù)，例：SKIPIF1<0；②拆：例：SKIPIF1<0；③除：例：SKIPIF1<0；④1的代入：例：已知SKIPIF1<0，求SKIPIF1<0的最小值.解析：SKIPIF1<0.⑤整體解：例：已知SKIPIF1<0,SKIPIF1<0是正數(shù)，且SKIPIF1<0，求SKIPIF1<0的最小值.解析：SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.2、對(duì)勾函數(shù)對(duì)勾函數(shù)是一種類(lèi)似于反比例函數(shù)的一般雙曲函數(shù)，是形如：SKIPIF1<0（SKIPIF1<0）的函數(shù).由圖象得名，又被稱為：“雙勾函數(shù)”、“對(duì)號(hào)函數(shù)”、“雙飛燕函數(shù)”、“耐克函數(shù)”等.函數(shù)SKIPIF1<0（SKIPIF1<0）常考對(duì)勾函數(shù)SKIPIF1<0（SKIPIF1<0）定義域SKIPIF1<0定義域SKIPIF1<0值域SKIPIF1<0值域SKIPIF1<0奇偶性奇函數(shù)奇偶性奇函數(shù)單調(diào)性SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0，SKIPIF1<0單調(diào)遞減單調(diào)性SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0，SKIPIF1<0單調(diào)遞減二、題型精講題型01直接法【典例1】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．4【答案】B【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，即函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，故選：B.【典例2】（2023春·安徽六安·高一?？计谥校┤鬝KIPIF1<0，則SKIPIF1<0（

）A．有最小值SKIPIF1<0 B．有最大值SKIPIF1<0C．有最小值2 D．有最大值2【答案】B【詳解】因?yàn)镾KIPIF1<0，則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，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).故選：B.【典例3】（2023·湖南長(zhǎng)沙·高二長(zhǎng)郡中學(xué)校考學(xué)業(yè)考試）代數(shù)式SKIPIF1<0取得最小值時(shí)對(duì)應(yīng)的SKIPIF1<0值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0在分母的位置，則SKIPIF1<0．SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)，取等號(hào),故選：D．【變式1】（2023秋·福建·高二統(tǒng)考學(xué)業(yè)考試）已知SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．2 B．3 C．4 D．5【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：C.【變式2】（2023春·福建福州·高二福建省福州延安中學(xué)?？紝W(xué)業(yè)考試）函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．2SKIPIF1<0 D．4【答案】D【詳解】∵SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故函數(shù)SKIPIF1<0的最小值為4.故選：D.題型02湊配法【典例1】（2023·高一課時(shí)練習(xí)）若SKIPIF1<0，則SKIPIF1<0的最值情況是（

）A．有最大值SKIPIF1<0 B．有最小值6 C．有最大值SKIPIF1<0 D．有最小值2【答案】B【詳解】若SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0等號(hào)成立，所以若SKIPIF1<0時(shí)，SKIPIF1<0有最小值為6，無(wú)最大值.故選：B.【典例2】（2023·安徽安慶·安慶一中校考三模）已知非負(fù)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值是___________.【答案】4【詳解】由SKIPIF1<0，可得SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào).故答案為：4【典例3】（2023·高一課時(shí)練習(xí)）當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，則a的取值范圍是__________．【答案】SKIPIF1<0【詳解】由SKIPIF1<0可得SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，因?yàn)镾KIPIF1<0恒成立，所以SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023春·內(nèi)蒙古呼倫貝爾·高一?？奸_(kāi)學(xué)考試）若SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值為（

）A．3 B．4 C．5 D．6【答案】D【詳解】由題意可得：SKIPIF1<0，∵SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.故選：D.【變式2】（2023春·山東德州·高二?？茧A段練習(xí)）已知正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________.【答案】SKIPIF1<0/1.125【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0最取到等號(hào).故答案為：SKIPIF1<0.題型03分離法【典例1】（2023春·江蘇泰州·高二泰州中學(xué)?？茧A段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的最小值是A．2 B．3 C．4 D．5【答案】D【詳解】由題意知，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取“=”）故SKIPIF1<0的最小值是5.故答案為D.【典例2】（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的最大值為_(kāi)_______.【答案】SKIPIF1<0/SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0≤SKIPIF1<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.【典例3】（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值是______．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.所以函數(shù)SKIPIF1<0的最小值是SKIPIF1<0故答案為:SKIPIF1<0.【變式1】（2022·江蘇·高一專題練習(xí)）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．4【答案】B【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．故選：B．【變式2】（2022秋·河北滄州·高一任丘市第一中學(xué)校考期中）解答下列問(wèn)題：(1)已知SKIPIF1<0，求函數(shù)SKIPIF1<0的最小值；(2)已知SKIPIF1<0，求函數(shù)SKIPIF1<0最小值．【答案】(1)10；(2)9.【詳解】（1）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，所以函數(shù)SKIPIF1<0的最小值為SKIPIF1<0（2）因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.所以函數(shù)SKIPIF1<0的最小值為9.題型04換元法【典例1】（2023春·重慶沙坪壩·高二重慶一中校考期中）已知SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0.于是SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0時(shí)，SKIPIF1<0取到最小值SKIPIF1<0.故選：C【典例2】（2023·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【詳解】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.【典例3】（2023·江蘇·高一專題練習(xí)）求下列函數(shù)的最小值（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【答案】（1）3；（2）SKIPIF1<0；（3）10.【詳解】（1）SKIPIF1<0∵SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0，即x=1時(shí)取“=”）即SKIPIF1<0的最小值為3；（2）令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0是單增，∴當(dāng)t=2時(shí)，y取最小值SKIPIF1<0；即y的最小值為SKIPIF1<0（3）令SKIPIF1<0，則SKIPIF1<0可化為：SKIPIF1<0當(dāng)且僅當(dāng)t=3時(shí)取“=”即y的最小值為10【變式1】（2023·全國(guó)·高三專題練習(xí)）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為_(kāi)_______.【答案】3【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，又由SKIPIF1<0得SKIPIF1<0，而函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，因此SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0，故答案為：SKIPIF1<0.【變式2】（2023·全國(guó)·高三專題練習(xí)）（1）求函數(shù)SKIPIF1<0的最小值及此時(shí)SKIPIF1<0的值；（2）已知函數(shù)SKIPIF1<0，SKIPIF1<0，求此函數(shù)的最小值及此時(shí)SKIPIF1<0的值.【答案】（1）函數(shù)SKIPIF1<0的最小值為5，此時(shí)SKIPIF1<0；（2）函數(shù)SKIPIF1<0的最小值為5，此時(shí)SKIPIF1<0.【詳解】（1）∵SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，等號(hào)成立.故函數(shù)SKIPIF1<0的最小值為5，此時(shí)SKIPIF1<0；（2）令SKIPIF1<0，將SKIPIF1<0代入得：SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.故函數(shù)SKIPIF1<0的最小值為5，此時(shí)SKIPIF1<0.題型05常數(shù)代換“1”的代換【典例1】（2023·高一單元測(cè)試）設(shè)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<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.【典例2】（2023·遼寧遼陽(yáng)·統(tǒng)考二模）若SKIPIF1<0，則SKIPIF1<0的值可以是__________．【答案】5（答案不唯一，只要不小于SKIPIF1<0即可）【詳解】因?yàn)镾KIPIF1<0，所以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，則SKIPIF1<0.【典例3】（2023秋·重慶長(zhǎng)壽·高一統(tǒng)考期末）已知正數(shù)SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________.【答案】SKIPIF1<0【詳解】由正數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，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.【典例4】（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_____．【答案】2【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.【變式1】（2023春·浙江·高二統(tǒng)考學(xué)業(yè)考試）正實(shí)數(shù)x，y滿足SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．3 B．7 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，由于SKIPIF1<0，由于SKIPIF1<0為正數(shù)，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故選：C【變式2】（2023·山東日照·三模）設(shè)SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)________．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，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，SKIPIF1<0時(shí)取得最小值．故答案為：SKIPIF1<0.故答案為：5（答案不唯一，只要不小于SKIPIF1<0即可）【變式3】（2023春·廣東汕頭·高一金山中學(xué)校考期中）已知正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________.【答案】SKIPIF1<0/SKIPIF1<0【詳解】因?yàn)檎龑?shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<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，故答案為：SKIPIF1<0【變式4】（2023春·吉林長(zhǎng)春·高二校考期中）已知正數(shù)SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)______.【答案】SKIPIF1<0【詳解】因?yàn)檎龜?shù)SKIPIF1<0、SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，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.題型06消元法【典例1】（2023·全國(guó)·高三專題練習(xí)）若正實(shí)數(shù)x，y滿足x＋2y＋xy＝7，則x＋y的最小值為（

）A．6 B．5 C．4 D．3【答案】D【詳解】因?yàn)閤＋2y＋xy＝7，所以SKIPIF1<0，所以SKIPIF1<0．因?yàn)镾KIPIF1<0，則SKIPIF1<0所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即x＝1，y＝2時(shí)，等號(hào)成立，所以x＋y的最小值為3．故選：D【典例2】（2023·全國(guó)·高三專題練習(xí)）已知正實(shí)數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)，取等號(hào)，所以SKIPIF1<0的最小值是SKIPIF1<0.故選：A.【變式1】（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．2SKIPIF1<0 D．2SKIPIF1<0【答案】D【詳解】由SKIPIF1<0，得SKIPIF1<0，而SKIPIF1<0，則有SKIPIF1<0，因此，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取“=”，所以SKIPIF1<0的最小值為2SKIPIF1<0.故選：D【變式2】（2023·全國(guó)·高三專題練習(xí)）已知正實(shí)數(shù)a，b滿足SKIPIF1<0，則SKIPIF1<0的最小值是（）A．2 B．SKIPIF1<0 C．SKIPIF1<0