人教A版高中數(shù)學(xué)（必修第一冊(cè)）同步講義 拓展 一元二次（分式）不等式解法及對(duì)勾函數(shù)（教師版）

第第頁(yè)拓展一一元二次（分式）不等式解法（含參數(shù)討論問(wèn)題）一、知識(shí)清單知識(shí)點(diǎn)01：一元二次不等式的有關(guān)概念1、一元二次不等式只含有一個(gè)未知數(shù)，并且未知數(shù)的最高次數(shù)是2的不等式，叫做一元二次不等式,一元二次不等式的一般形式：①SKIPIF1<0（其中SKIPIF1<0均為常數(shù)）②SKIPIF1<0（其中SKIPIF1<0均為常數(shù)）③SKIPIF1<0（其中SKIPIF1<0均為常數(shù)）④SKIPIF1<0（其中SKIPIF1<0均為常數(shù)）2、一元二次不等式的解與解集使某一個(gè)一元二次不等式成立的SKIPIF1<0的值，叫作這個(gè)一元二次不等式的解，其解的集合，稱(chēng)為這個(gè)一元二次不等式的解集.將一個(gè)不等式轉(zhuǎn)化為另一個(gè)與它解集相同的不等式，叫作不等式的同解變形.知識(shí)點(diǎn)02：四個(gè)二次的關(guān)系1一元二次函數(shù)的零點(diǎn)一般地，對(duì)于二次函數(shù)SKIPIF1<0，我們把使SKIPIF1<0的實(shí)數(shù)SKIPIF1<0叫做二次函數(shù)SKIPIF1<0的零點(diǎn)．2次函數(shù)與一元二次方程的根、一元二次不等式的解集的對(duì)應(yīng)關(guān)系對(duì)于一元二次方程SKIPIF1<0的兩根為SKIPIF1<0且SKIPIF1<0，設(shè)SKIPIF1<0，它的解按照SKIPIF1<0，SKIPIF1<0，SKIPIF1<0可分三種情況，相應(yīng)地，二次函數(shù)SKIPIF1<0SKIPIF1<0的圖象與SKIPIF1<0軸的位置關(guān)系也分為三種情況.因此我們分三種情況來(lái)討論一元二次不等式SKIPIF1<0SKIPIF1<0或SKIPIF1<0SKIPIF1<0的解集.判別式SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0二次函數(shù)SKIPIF1<0(SKIPIF1<0的圖象一元二次方程SKIPIF1<0(SKIPIF1<0)的根有兩個(gè)不相等的實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0(SKIPIF1<0)有兩個(gè)相等的實(shí)數(shù)根SKIPIF1<0沒(méi)有實(shí)數(shù)根SKIPIF1<0(SKIPIF1<0)的解集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(SKIPIF1<0)的解集SKIPIF1<0SKIPIF1<0SKIPIF1<0知識(shí)點(diǎn)03：一元二次不等式的解法1：先看二次項(xiàng)系數(shù)是否為正，若為負(fù)，則將二次項(xiàng)系數(shù)化為正數(shù)；2：寫(xiě)出相應(yīng)的方程SKIPIF1<0SKIPIF1<0，計(jì)算判別式SKIPIF1<0：①SKIPIF1<0時(shí)，求出兩根SKIPIF1<0，且SKIPIF1<0（注意靈活運(yùn)用十字相乘法）；②SKIPIF1<0時(shí)，求根SKIPIF1<0；③SKIPIF1<0時(shí)，方程無(wú)解3：根據(jù)不等式，寫(xiě)出解集.知識(shí)點(diǎn)04：解分式不等式1、分式不等式定義：與分式方程類(lèi)似，分母中含有未知數(shù)的不等式稱(chēng)為分式不等式，如：形如SKIPIF1<0或SKIPIF1<0（其中SKIPIF1<0,SKIPIF1<0為整式且SKIPIF1<0的不等式稱(chēng)為分式不等式。2、分式不等式的解法①移項(xiàng)化零：將分式不等式右邊化為0：②SKIPIF1<0③SKIPIF1<0④SKIPIF1<0⑤SKIPIF1<0二、題型精講題型01一元二次不等式（不含參）的求解（首項(xiàng)系數(shù)為正）【典例1】（2023·上海金山·統(tǒng)考二模）若實(shí)數(shù)SKIPIF1<0滿(mǎn)足不等式SKIPIF1<0，則SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【詳解】不等式SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.【典例2】（2023·江西九江·校考模擬預(yù)測(cè)）SKIPIF1<0的解集是_______.【答案】SKIPIF1<0【詳解】由SKIPIF1<0，解得：SKIPIF1<0.所以SKIPIF1<0的解集是：SKIPIF1<0故答案為：SKIPIF1<0【典例3】（2023·高一課時(shí)練習(xí)）不等式SKIPIF1<0的解集為_(kāi)_____．【答案】SKIPIF1<0【詳解】解：因?yàn)椴坏仁絊KIPIF1<0可化為SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以原不等式的解集為SKIPIF1<0故答案為：SKIPIF1<0【變式1】（2023·上海長(zhǎng)寧·統(tǒng)考一模）不等式SKIPIF1<0的解集為_(kāi)__________【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以不等式的解集為：SKIPIF1<0，故答案為：SKIPIF1<0.【變式2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）不等式SKIPIF1<0的解集為_(kāi)_____.【答案】SKIPIF1<0【詳解】由SKIPIF1<0得：SKIPIF1<0

SKIPIF1<0即不等式SKIPIF1<0的解集為SKIPIF1<0故答案為：SKIPIF1<0題型02一元二次不等式（不含參）的求解（首項(xiàng)系數(shù)為負(fù)）【典例1】（2023春·山東濱州·高一?？茧A段練習(xí)）不等式SKIPIF1<0的解集為_(kāi)__________【答案】SKIPIF1<0【詳解】SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0所以不等式的解集為SKIPIF1<0.故答案為：SKIPIF1<0【典例2】（2023秋·北京順義·高一統(tǒng)考期末）不等式SKIPIF1<0的解集是__________.【答案】SKIPIF1<0或SKIPIF1<0【詳解】解：不等式SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式的解集為SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0【典例3】（2023秋·黑龍江七臺(tái)河·高三?？计谥校┎坏仁絊KIPIF1<0的解集為_(kāi)__________.【答案】SKIPIF1<0【詳解】由SKIPIF1<0等價(jià)于SKIPIF1<0，可得SKIPIF1<0.故答案為：SKIPIF1<0【變式1】（2023·上?！じ呷y(tǒng)考學(xué)業(yè)考試）一元二次不等式SKIPIF1<0的解集為_(kāi)_____________【答案】SKIPIF1<0或SKIPIF1<0【詳解】由SKIPIF1<0，得SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式的解集為SKIPIF1<0或SKIPIF1<0；故答案為：SKIPIF1<0或SKIPIF1<0【變式2】（2023·江西九江·?？寄M預(yù)測(cè)）不等式SKIPIF1<0的解集為_(kāi)_____．【答案】SKIPIF1<0【詳解】SKIPIF1<0，因?yàn)橐辉畏匠蘏KIPIF1<0的判別式SKIPIF1<0，二次函數(shù)SKIPIF1<0的開(kāi)口向上，所以不等式SKIPIF1<0的解集為空集，故答案為：SKIPIF1<0【變式3】（2023秋·天津南開(kāi)·高一天津大學(xué)附屬中學(xué)?？计谀┎坏仁絊KIPIF1<0的解為_(kāi)__________.【答案】SKIPIF1<0【詳解】由題意，SKIPIF1<0，即求解不等式SKIPIF1<0，解得SKIPIF1<0，所以不等式的解集為SKIPIF1<0.故答案為：SKIPIF1<0題型03分式不等式【典例1】（2023·上海寶山·上海交大附中?？既＃┎坏仁絊KIPIF1<0的解集為_(kāi)_________.【答案】SKIPIF1<0【詳解】原不等式等價(jià)于SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0,故答案為：SKIPIF1<0.【典例2】（2023秋·上海徐匯·高一統(tǒng)考期末）不等式SKIPIF1<0的解集為_(kāi)_______．【答案】SKIPIF1<0【詳解】由于SKIPIF1<0，所以不等式SKIPIF1<0即不等式SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0，故答案為：SKIPIF1<0【典例3】（2023春·上海長(zhǎng)寧·高三上海市延安中學(xué)?？奸_(kāi)學(xué)考試）不等式SKIPIF1<0的解集是________．【答案】SKIPIF1<0或SKIPIF1<0}【詳解】SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式SKIPIF1<0的解集是SKIPIF1<0或SKIPIF1<0}.故答案為：SKIPIF1<0或SKIPIF1<0}【典例4】（2023春·上海徐匯·高三上海民辦南模中學(xué)?？茧A段練習(xí)）不等式SKIPIF1<0的解集是__________.【答案】SKIPIF1<0或SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，滿(mǎn)足條件的SKIPIF1<0不存在，所以不等式SKIPIF1<0的解集是SKIPIF1<0或SKIPIF1<0，故答案為：SKIPIF1<0或SKIPIF1<0.【變式1】（2023春·上海普陀·高三曹楊二中?？茧A段練習(xí)）不等式SKIPIF1<0的解集是___________.【答案】SKIPIF1<0【詳解】不等式SKIPIF1<0即SKIPIF1<0，故不等式SKIPIF1<0的解集是SKIPIF1<0，故答案為：SKIPIF1<0【變式2】（2023春·上海嘉定·高三統(tǒng)考階段練習(xí)）不等式SKIPIF1<0的解集為_(kāi)_____.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以不等式的解集為SKIPIF1<0故答案為:SKIPIF1<0【變式3】（2023秋·陜西渭南·高二統(tǒng)考期末）（1）解不等式SKIPIF1<0；【答案】（1）SKIPIF1<0或SKIPIF1<0；【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以原不等式的解集為SKIPIF1<0或SKIPIF1<0.題型04一元二次不等式（含參）的求解（兩根大小不確定從兩根相等開(kāi)始討論）【典例1】（2023秋·遼寧沈陽(yáng)·高一統(tǒng)考期末）求關(guān)于x的不等式SKIPIF1<0的解集．【答案】答案見(jiàn)解析【詳解】原不等式可化為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0．【典例2】（2023·全國(guó)·高一專(zhuān)題練習(xí)）解下列關(guān)于SKIPIF1<0的不等式：SKIPIF1<0．【答案】答案見(jiàn)解析.【詳解】由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，解集為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，解集為SKIPIF1<0．綜上：SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；SKIPIF1<0時(shí)，解集為SKIPIF1<0；SKIPIF1<0時(shí)，解集為SKIPIF1<0．【典例3】（2023秋·山東泰安·高一統(tǒng)考期末）已知關(guān)于x的不等式SKIPIF1<0．(1)若不等式的解集為SKIPIF1<0，求a，b的值：(2)若SKIPIF1<0，解不等式SKIPIF1<0．【答案】(1)SKIPIF1<0(2)答案見(jiàn)解析．【詳解】（1）原不等式可化為SKIPIF1<0，由題知，SKIPIF1<0，SKIPIF1<0是方程SKIPIF1<0的兩根，由根與系數(shù)的關(guān)系得SKIPIF1<0，解得SKIPIF1<0．（2）原不等式可化為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以原不等式化為SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，解得SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，解得SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，解得SKIPIF1<0；綜上所述，當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0．【典例4】（2023秋·山東淄博·高一統(tǒng)考期末）已知一元二次函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求實(shí)數(shù)a的取值范圍；(2)求關(guān)于x的不等式SKIPIF1<0的解集．【答案】(1)SKIPIF1<0(2)答案見(jiàn)解析【詳解】（1）由已知得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.實(shí)數(shù)a的取值范圍SKIPIF1<0；（2）SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的解集為SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的解集為SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的解集為SKIPIF1<0，綜上所述：當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0；【變式1】（2023·高一單元測(cè)試）已知關(guān)于x的不等式SKIPIF1<0的解集為SKIPIF1<0．(1)求實(shí)數(shù)a，b的值；(2)解關(guān)于x的不等式SKIPIF1<0．【答案】(1)SKIPIF1<0(2)見(jiàn)解析【詳解】（1）因?yàn)椴坏仁絊KIPIF1<0的解集為SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0是方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根，SKIPIF1<0且SKIPIF1<0．由根與系數(shù)的關(guān)系，得SKIPIF1<0,解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0.（2）由（1）知，不等式SKIPIF1<0可化為SKIPIF1<0，即SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0的解集為SKIPIF1<0．綜上所述：當(dāng)SKIPIF1<0時(shí)，原不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，原不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，原不等式的解集為SKIPIF1<0．【變式2】（2023春·遼寧沈陽(yáng)·高二新民市高級(jí)中學(xué)?？茧A段練習(xí)）已知不等式SKIPIF1<0的解集為SKIPIF1<0．(1)求實(shí)數(shù)a，b的值；(2)解不等式SKIPIF1<0．【答案】(1)SKIPIF1<0;(2)答案見(jiàn)解析【詳解】（1）SKIPIF1<0的解集為SKIPIF1<0，SKIPIF1<0和SKIPIF1<0是SKIPIF1<0的兩個(gè)根，根據(jù)根與系數(shù)的關(guān)系可知：SKIPIF1<0，SKIPIF1<0；（2）由（1）可知SKIPIF1<0，SKIPIF1<0即SKIPIF1<0,SKIPIF1<0,①當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)解集為SKIPIF1<0且SKIPIF1<0；②當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，此時(shí)解集為SKIPIF1<0或SKIPIF1<0；③當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，此時(shí)解集為SKIPIF1<0或SKIPIF1<0；綜上：當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0且SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，解集為SKIPIF1<0或SKIPIF1<0題型05一元二次不等式（含參）的求解（首項(xiàng)系數(shù)含參從0開(kāi)始討論）【典例1】（2023春·四川瀘州·高二?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，解不等式SKIPIF1<0.【答案】答案見(jiàn)解析【詳解】①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；∴SKIPIF1<0.②當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，（i）當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，（ⅱ）當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，（ⅲ）當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，綜上，當(dāng)SKIPIF1<0時(shí)，所求不等式的解集為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，所求不等式的解集為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，所求不等式的解集為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，所求不等式的解集為SKIPIF1<0.【典例2】（2023秋·重慶江北·高一字水中學(xué)?？计谀?）若不等式SKIPIF1<0對(duì)一切實(shí)數(shù)SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍；（2）解關(guān)于SKIPIF1<0的不等式SKIPIF1<0．【答案】（1）SKIPIF1<0；（2）答案見(jiàn)解析.【詳解】（1）由題意，SKIPIF1<0恒成立，當(dāng)SKIPIF1<0時(shí)，不等式可化為SKIPIF1<0，不滿(mǎn)足題意；當(dāng)SKIPIF1<0時(shí)，滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0；故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．（2）不等式SKIPIF1<0等價(jià)于SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，不等式可化為SKIPIF1<0，所以不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式可化為SKIPIF1<0，此時(shí)SKIPIF1<0，所以不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式可化為SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式的解集為SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式的解集為SKIPIF1<0或SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不等式的解集為SKIPIF1<0或SKIPIF1<0.綜上：SKIPIF1<0時(shí)，等式的解集為SKIPIF1<0或SKIPIF1<0SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0或SKIPIF1<0SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0【典例3】（2023春·湖北武漢·高一華中師大一附中?？茧A段練習(xí)）已知SKIPIF1<0，解關(guān)于SKIPIF1<0的不等式SKIPIF1<0．【答案】答案見(jiàn)解析【詳解】當(dāng)SKIPIF1<0時(shí)，不等式為SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式為SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式為SKIPIF1<0，若SKIPIF1<0，不等式為SKIPIF1<0，解得SKIPIF1<0；若SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.綜上所述，當(dāng)SKIPIF1<0時(shí)，原不等式的解集是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，原不等式的解集是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，原不等式的解集是SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，原不等式的解集是SKIPIF1<0或SKIPIF1<0．【典例4】（2023秋·河北邯鄲·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，解不等式SKIPIF1<0；(2)解關(guān)于SKIPIF1<0的不等式SKIPIF1<0.【答案】(1)SKIPIF1<0或SKIPIF1<0(2)答案見(jiàn)解析【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以由SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0.（2）由SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，解得SKIPIF1<0，故不等式的解集為SKIPIF1<0；令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，不等式解得SKIPIF1<0或SKIPIF1<0，故不等式的解集為SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，解得SKIPIF1<0，故不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，不等式解得SKIPIF1<0或SKIPIF1<0，故不等式的解集為SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，不等式解得SKIPIF1<0，故不等式的解集為SKIPIF1<0；綜上：當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0或SKIPIF1<0；【變式1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知二次函數(shù)SKIPIF1<0．(1)若SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．(2)解關(guān)于SKIPIF1<0的不等式SKIPIF1<0（其中SKIPIF1<0）．【答案】(1)SKIPIF1<0；(2)答案見(jiàn)解析.【詳解】（1）不等式SKIPIF1<0即為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式可變形為：SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，所以實(shí)數(shù)a的取值范圍是SKIPIF1<0.（2）不等式SKIPIF1<0，等價(jià)于SKIPIF1<0，即SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，不等式整理為SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0的兩根為SKIPIF1<0，SKIPIF1<0，②當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，解不等式SKIPIF1<0得SKIPIF1<0或SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，解不等式SKIPIF1<0得SKIPIF1<0；④當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，不等式SKIPIF1<0的解集為SKIPIF1<0；⑤當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，解不等式SKIPIF1<0得SKIPIF1<0；綜上所述，不等式的解集為：①當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；④當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；⑤當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0．【變式2】（2023秋·山東濟(jì)寧·高一濟(jì)寧一中?？计谀┮阎坏仁絊KIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0，(1)求a，b的值；(2)解關(guān)于x的不等式SKIPIF1<0．【答案】(1)SKIPIF1<0(2)答案見(jiàn)解析【詳解】（1）由題意知一元二次方程SKIPIF1<0的解為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，由韋達(dá)定理有：SKIPIF1<0.（2）由（1）知SKIPIF1<0，則原不等式等價(jià)于SKIPIF1<0，因式分解得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)：不等式的解集為：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)：不等式的解集為：SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)：不等式的解集為：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)：不等式的解集為：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)：不等式的解集為：SKIPIF1<0；【變式3】（2023·全國(guó)·高三專(zhuān)題練習(xí)）解關(guān)于x的不等式SKIPIF1<0．【答案】詳見(jiàn)解析.【詳解】原不等式變形為SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，不等式即為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，xSKIPIF1<0或SKIPIF1<0；由于SKIPIF1<0，于是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．綜上，當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式的解集為SKIPIF1<0．題型06一元二次不等式（含參）的求解（不可因式分解型）【典例1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）解關(guān)于x的不等式SKIPIF1<0．【答案】答案見(jiàn)解析.【詳解】解：（1）當(dāng)SKIPIF1<0時(shí)，原不等式SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0不等式解集為SKIPIF1<0；（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0開(kāi)口向上，由圖象得：SKIPIF1<0若SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0的解集為SKIPIF1<0；SKIPIF1<0若SKIPIF1<0時(shí)，SKIPIF1<0，不等式SKIPIF1<0解集為SKIPIF1<0；（3）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0，SKIPIF1<0SKIPIF1<0開(kāi)口向下，由圖象得不等式解集為SKIPIF1<0；綜上可知，當(dāng)SKIPIF1<0時(shí)不等式解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0．【典例2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）解下列關(guān)于SKIPIF1<0的不等式：SKIPIF1<0（SKIPIF1<0）；【答案】（1）SKIPIF1<0；（2）見(jiàn)詳解；（3）見(jiàn)詳解.【詳解】）SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0無(wú)實(shí)數(shù)解，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的無(wú)實(shí)數(shù)解，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的解為SKIPIF1<0，綜上，當(dāng)SKIPIF1<0時(shí)，原不等式的解集為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，原不等式的解集為SKIPIF1<0【變式1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）解下列關(guān)于SKIPIF1<0的不等式SKIPIF1<0．【答案】答案見(jiàn)解析【詳解】由對(duì)應(yīng)函數(shù)SKIPIF1<0開(kāi)口向上，且SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0恒成立，原不等式解集為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0時(shí)，由SKIPIF1<0，可得SKIPIF1<0，所以原不等式解集為SKIPIF1<0；綜上，SKIPIF1<0解集為SKIPIF1<0；SKIPIF1<0或SKIPIF1<0解集為SKIPIF1<0.

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(chēng)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）?？紝?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·高一校考期中）若SKIPIF1<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)?？紝W(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)校考學(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(mǎn)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·高一校考開(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·高二校考階段練習(xí)）已知正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_________.【答案】SKIPIF1<0/1.125【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)校考階段練習(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(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的最大值為_(kāi)_______.【答案】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(zhuān)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(zhuān)題練習(xí)）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．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)?？计谥校┙獯鹣铝袉?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