人教A版高中數(shù)學（必修第一冊）同步講義 拓展 一元二次（分式）不等式解法及對勾函數(shù)（原卷版）

第第頁拓展一一元二次（分式）不等式解法（含參數(shù)討論問題）一、知識清單知識點01：一元二次不等式的有關概念1、一元二次不等式只含有一個未知數(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一元二次不等式成立的SKIPIF1<0的值，叫作這個一元二次不等式的解，其解的集合，稱為這個一元二次不等式的解集.將一個不等式轉化為另一個與它解集相同的不等式，叫作不等式的同解變形.知識點02：四個二次的關系1一元二次函數(shù)的零點一般地，對于二次函數(shù)SKIPIF1<0，我們把使SKIPIF1<0的實數(shù)SKIPIF1<0叫做二次函數(shù)SKIPIF1<0的零點．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應關系對于一元二次方程SKIPIF1<0的兩根為SKIPIF1<0且SKIPIF1<0，設SKIPIF1<0，它的解按照SKIPIF1<0，SKIPIF1<0，SKIPIF1<0可分三種情況，相應地，二次函數(shù)SKIPIF1<0SKIPIF1<0的圖象與SKIPIF1<0軸的位置關系也分為三種情況.因此我們分三種情況來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)的根有兩個不相等的實數(shù)根SKIPIF1<0，SKIPIF1<0(SKIPIF1<0)有兩個相等的實數(shù)根SKIPIF1<0沒有實數(shù)根SKIPIF1<0(SKIPIF1<0)的解集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(SKIPIF1<0)的解集SKIPIF1<0SKIPIF1<0SKIPIF1<0知識點03：一元二次不等式的解法1：先看二次項系數(shù)是否為正，若為負，則將二次項系數(shù)化為正數(shù)；2：寫出相應的方程SKIPIF1<0SKIPIF1<0，計算判別式SKIPIF1<0：①SKIPIF1<0時，求出兩根SKIPIF1<0，且SKIPIF1<0（注意靈活運用十字相乘法）；②SKIPIF1<0時，求根SKIPIF1<0；③SKIPIF1<0時，方程無解3：根據(jù)不等式，寫出解集.知識點04：解分式不等式1、分式不等式定義：與分式方程類似，分母中含有未知數(shù)的不等式稱為分式不等式，如：形如SKIPIF1<0或SKIPIF1<0（其中SKIPIF1<0,SKIPIF1<0為整式且SKIPIF1<0的不等式稱為分式不等式。2、分式不等式的解法①移項化零：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（首項系數(shù)為正）【典例1】若實數(shù)SKIPIF1<0滿足不等式SKIPIF1<0，則SKIPIF1<0的取值范圍是__________.【典例2】SKIPIF1<0的解集是_______.【典例3】不等式SKIPIF1<0的解集為______．【變式1】不等式SKIPIF1<0的解集為___________【變式2】不等式SKIPIF1<0的解集為______.題型02一元二次不等式（不含參）的求解（首項系數(shù)為負）【典例1】不等式SKIPIF1<0的解集為___________【典例2】不等式SKIPIF1<0的解集是__________.【典例3】不等式SKIPIF1<0的解集為___________.【變式1】一元二次不等式SKIPIF1<0的解集為______________【變式2】不等式SKIPIF1<0的解集為______．【變式3】不等式SKIPIF1<0的解為___________.題型03分式不等式【典例1】不等式SKIPIF1<0的解集為__________.【典例2】不等式SKIPIF1<0的解集為________．【典例3】不等式SKIPIF1<0的解集是________．【典例4】不等式SKIPIF1<0的解集是__________.【變式1】不等式SKIPIF1<0的解集是___________.【變式2】不等式SKIPIF1<0的解集為______.【變式3】解不等式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始討論）【典例1】求關于x的不等式SKIPIF1<0的解集．【典例2】解下列關于SKIPIF1<0的不等式：SKIPIF1<0．【典例3】已知關于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．【典例4】已知一元二次函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求實數(shù)a的取值范圍；(2)求關于x的不等式SKIPIF1<0的解集．【變式1】已知關于x的不等式SKIPIF1<0的解集為SKIPIF1<0．(1)求實數(shù)a，b的值；(2)解關于x的不等式SKIPIF1<0．【變式2】已知不等式SKIPIF1<0的解集為SKIPIF1<0．(1)求實數(shù)a，b的值；(2)解不等式SKIPIF1<0．題型05一元二次不等式（含參）的求解（首項系數(shù)含參從0開始討論）【典例1】已知函數(shù)SKIPIF1<0，解不等式SKIPIF1<0.【典例2】（1）若不等式SKIPIF1<0對一切實數(shù)SKIPIF1<0恒成立，求實數(shù)SKIPIF1<0的取值范圍；（2）解關于SKIPIF1<0的不等式SKIPIF1<0．【典例3】已知SKIPIF1<0，解關于SKIPIF1<0的不等式SKIPIF1<0．【典例4】已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，解不等式SKIPIF1<0；(2)解關于SKIPIF1<0的不等式SKIPIF1<0.【變式1】已知二次函數(shù)SKIPIF1<0．(1)若SKIPIF1<0時，不等式SKIPIF1<0恒成立，求實數(shù)SKIPIF1<0的取值范圍．(2)解關于SKIPIF1<0的不等式SKIPIF1<0（其中SKIPIF1<0）．【變式2】已知不等式SKIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0，(1)求a，b的值；(2)解關于x的不等式SKIPIF1<0．【變式3】解關于x的不等式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】解關于x的不等式SKIPIF1<0．【典例2】解下列關于SKIPIF1<0的不等式：SKIPIF1<0（SKIPIF1<0）；【變式1】（2023·全國·高三專題練習）解下列關于SKIPIF1<0的不等式SKIPIF1<0．

第07講拓展二基本不等式與對勾函數(shù)一、知識清單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，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、對勾函數(shù)對勾函數(shù)是一種類似于反比例函數(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)”、“對號函數(shù)”、“雙飛燕函數(shù)”、“耐克函數(shù)”等.函數(shù)SKIPIF1<0（SKIPIF1<0）?？紝春瘮?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】函數(shù)SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．4【典例2】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【典例3】代數(shù)式SKIPIF1<0取得最小值時對應的SKIPIF1<0值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1】已知SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．2 B．3 C．4 D．5【變式2】函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．2SKIPIF1<0 D．4題型02湊配法【典例1】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【典例2】已知非負數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值是___________.【典例3】當SKIPIF1<0時，不等式SKIPIF1<0恒成立，則a的取值范圍是__________．【變式1】若SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值為（

）A．3 B．4 C．5 D．6【變式2】已知正實數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為__________.題型03分離法【典例1】已知SKIPIF1<0，則SKIPIF1<0的最小值是A．2 B．3 C．4 D．5【典例2】函數(shù)SKIPIF1<0的最大值為________.【典例3】已知SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值是______．【變式1】當SKIPIF1<0時，函數(shù)SKIPIF1<0的最小值為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．4【變式2】解答下列問題：(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．題型04換元法【典例1】已知SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【典例2】若實數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為________.【典例3】求下列函數(shù)的最小值（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【變式1】當SKIPIF1<0時，SKIPIF1<0的最小值為________.【變式2】（1）求函數(shù)SKIPIF1<0的最小值及此時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SKIPIF1<0的值.題型05常數(shù)代換“1”的代換【典例1】設SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為__________．【典例2】若SKIPIF1<0，則SKIPIF1<0的值可以是__________．【典例3】已知正數(shù)SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的最小值為__________.【典例4】已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為______．【變式1】正實數(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【變式2】設SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值為_________．【變式3】已知正實數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為__________.【變式4】已知正數(shù)S