人教A版高中數(shù)學(xué)（必修第一冊(cè)）同步講義 4.1-4.2指數(shù)與指數(shù)函數(shù)（教師版）

第第頁(yè)4.1指數(shù)課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解根式和分?jǐn)?shù)指數(shù)冪的含義，并且能進(jìn)行兩者之間的互化。②掌握根式的性質(zhì)，并能運(yùn)用根式的運(yùn)算性質(zhì)進(jìn)行根式的運(yùn)算。③掌握實(shí)數(shù)指數(shù)冪的運(yùn)算性質(zhì)，學(xué)會(huì)化簡(jiǎn)較復(fù)雜的運(yùn)算式子。通過(guò)本節(jié)課的學(xué)習(xí)，能將初中的根式與本節(jié)課根式進(jìn)行順利對(duì)接與延伸，條件的擴(kuò)充使指數(shù)的運(yùn)算性質(zhì)內(nèi)容更充實(shí)，條件更充分，運(yùn)算更徹底，因此本節(jié)課的內(nèi)容具有承上啟下的作用，通過(guò)本節(jié)課的學(xué)習(xí)要求掌握根式和分?jǐn)?shù)指數(shù)冪的具體運(yùn)算，并能進(jìn)行兩者的互化，運(yùn)用實(shí)數(shù)指數(shù)冪的運(yùn)算性質(zhì)進(jìn)行化簡(jiǎn).知識(shí)點(diǎn)01：整數(shù)指數(shù)冪1、正整數(shù)指數(shù)冪的定義：SKIPIF1<0，其中，SKIPIF1<02、正整數(shù)指數(shù)冪的運(yùn)算法則：①SKIPIF1<0（SKIPIF1<0）②SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）③SKIPIF1<0（SKIPIF1<0）④SKIPIF1<0（SKIPIF1<0）⑤SKIPIF1<0（SKIPIF1<0SKIPIF1<0）知識(shí)點(diǎn)02：根式1、SKIPIF1<0次根式定義：一般地，如果SKIPIF1<0，那么SKIPIF1<0叫做SKIPIF1<0的SKIPIF1<0次方根，其中SKIPIF1<0，且SKIPIF1<0.特別的：①當(dāng)SKIPIF1<0是奇數(shù)時(shí)，正數(shù)的SKIPIF1<0次方根是一個(gè)正數(shù)，負(fù)數(shù)的SKIPIF1<0次方根是一個(gè)負(fù)數(shù).這時(shí)，SKIPIF1<0的SKIPIF1<0次方根用符號(hào)表示SKIPIF1<0.②當(dāng)SKIPIF1<0是偶數(shù)時(shí)，正數(shù)的SKIPIF1<0次方根有兩個(gè)，這兩個(gè)數(shù)互為相反數(shù).這時(shí)，正數(shù)SKIPIF1<0的正的SKIPIF1<0次方根用符號(hào)SKIPIF1<0表示，叫做SKIPIF1<0的SKIPIF1<0次算術(shù)根；負(fù)的SKIPIF1<0次方根用符號(hào)SKIPIF1<0表示.正的SKIPIF1<0次方根與負(fù)的SKIPIF1<0次方根可以合并寫成SKIPIF1<0(SKIPIF1<0).③負(fù)數(shù)沒有偶次方根；④SKIPIF1<0的任何次方根都是SKIPIF1<0，記作SKIPIF1<02、根式：式子SKIPIF1<0叫做根式，這里SKIPIF1<0叫做根指數(shù)，SKIPIF1<0叫做被開方數(shù)．在根式符號(hào)SKIPIF1<0中，注意：①SKIPIF1<0，SKIPIF1<0②當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0對(duì)任意SKIPIF1<0都有意義③當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0只有當(dāng)SKIPIF1<0時(shí)才有意義.3、SKIPIF1<0與SKIPIF1<0的區(qū)別：①當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0（SKIPIF1<0）②當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0（SKIPIF1<0）③當(dāng)SKIPIF1<0為奇數(shù)時(shí)，且SKIPIF1<0，SKIPIF1<0④SKIPIF1<0為偶數(shù)時(shí)，且SKIPIF1<0，SKIPIF1<0【即學(xué)即練1】（2023·全國(guó)·高一假期作業(yè)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0【答案】D【詳解】當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0.故選：D知識(shí)點(diǎn)03：分式指數(shù)冪1、正數(shù)的正分?jǐn)?shù)指數(shù)冪的意義是SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）于是，在條件SKIPIF1<0，SKIPIF1<0，SKIPIF1<0下，根式都可以寫成分?jǐn)?shù)指數(shù)冪的形式.2、正數(shù)的負(fù)分?jǐn)?shù)指數(shù)冪的意義與負(fù)整數(shù)指數(shù)冪的意義相仿，我們規(guī)定，SKIPIF1<0（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）.3、SKIPIF1<0的正分?jǐn)?shù)指數(shù)冪等于SKIPIF1<0，SKIPIF1<0的負(fù)分?jǐn)?shù)指數(shù)冪沒有意義.【即學(xué)即練2】（2023·全國(guó)·高三專題練習(xí)）化簡(jiǎn)：SKIPIF1<0=．（用分?jǐn)?shù)指數(shù)冪表示）.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.知識(shí)點(diǎn)04：有理數(shù)指數(shù)冪①SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）②SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）③SKIPIF1<0（SKIPIF1<0，SKIPIF1<0SKIPIF1<0）知識(shí)點(diǎn)05：無(wú)理數(shù)指數(shù)冪①SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）②SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）③SKIPIF1<0（SKIPIF1<0，SKIPIF1<0SKIPIF1<0）【即學(xué)即練3】（2023·高一課時(shí)練習(xí)）計(jì)算SKIPIF1<0.【答案】0【詳解】原式SKIPIF1<0SKIPIF1<0SKIPIF1<0題型01根式的概念【典例1】（2023·全國(guó)·高一假期作業(yè)）二次根式SKIPIF1<0成立的條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0是任意實(shí)數(shù)【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.故選：C.【典例2】（2023·高一課時(shí)練習(xí)）81的4次方根是．【答案】SKIPIF1<0【詳解】81的4次方根是SKIPIF1<0.故答案為：SKIPIF1<0【變式1】（2023·高一課時(shí)練習(xí)）625的四次方根為.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以625的四次方根為SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023·江蘇·高一假期作業(yè)）SKIPIF1<0的平方根為，SKIPIF1<0的SKIPIF1<0次方根為；已知SKIPIF1<0，則SKIPIF1<0；【答案】SKIPIF1<0SKIPIF1<0SKIPIF1<0【詳解】SKIPIF1<0，SKIPIF1<0的平方根為SKIPIF1<0.SKIPIF1<0的SKIPIF1<0次方根為SKIPIF1<0.SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0；SKIPIF1<0.題型02根式的化簡(jiǎn)(求值)【典例1】（2023·江蘇·高一假期作業(yè)）若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為（

）A．1 B．5 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】依題意，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的值為1.故選：A【典例2】（2023·江蘇·高一假期作業(yè)）當(dāng)SKIPIF1<0有意義時(shí)，化簡(jiǎn)SKIPIF1<0的結(jié)果是（

）A．2x－5B．－2x－1C．－1D．5－2x【答案】C【詳解】因?yàn)镾KIPIF1<0有意義，可得SKIPIF1<0，即SKIPIF1<0，又由SKIPIF1<0SKIPIF1<0故選：C.【變式1】（2023·江蘇·高一假期作業(yè)）若SKIPIF1<0，化簡(jiǎn)SKIPIF1<0的結(jié)果是（

）A．5－2a B．2a－5C．1 D．－1【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：C.【變式2】（2023·江蘇·高一假期作業(yè)）有下列說(shuō)法：①SKIPIF1<0；②16的4次方根是SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0．其中，正確的有(填序號(hào))．【答案】②④【詳解】n為奇數(shù)時(shí)，負(fù)數(shù)的n次方根是一個(gè)負(fù)數(shù)，SKIPIF1<0，故①錯(cuò)誤；16的4次方根有兩個(gè)，為SKIPIF1<0，故②正確；因?yàn)镾KIPIF1<0，故③錯(cuò)誤；因?yàn)镾KIPIF1<0是正數(shù)，故SKIPIF1<0，故④正確．故答案為：②④題型03分?jǐn)?shù)指數(shù)冪的簡(jiǎn)單計(jì)算【典例1】（多選）（2023·江蘇·高一假期作業(yè)）（多選題）下列各式中一定成立的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【詳解】SKIPIF1<0，SKIPIF1<0錯(cuò)誤；SKIPIF1<0，SKIPIF1<0正確；SKIPIF1<0，SKIPIF1<0錯(cuò)誤；SKIPIF1<0，SKIPIF1<0正確，故選：SKIPIF1<0【典例2】（2023·高一課時(shí)練習(xí)）根式SKIPIF1<0的分?jǐn)?shù)指數(shù)冪的形式為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0.故選D.【典例3】（2023秋·山西·高一校聯(lián)考期中）（1）化簡(jiǎn)：SKIPIF1<0.（結(jié)果用分?jǐn)?shù)指數(shù)冪表示）（2）化簡(jiǎn)：SKIPIF1<0.（結(jié)果用分?jǐn)?shù)指數(shù)冪表示）（3）求值：SKIPIF1<0.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【詳解】（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【變式1】（2023·高一單元測(cè)試）下列式子的互化正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】根據(jù)分?jǐn)?shù)指數(shù)冪的運(yùn)算可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故選：C【變式2】（2023·全國(guó)·高一假期作業(yè)）化簡(jiǎn)求值:(1)SKIPIF1<0；(2)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0SKIPIF1<0.（2）SKIPIF1<0SKIPIF1<0SKIPIF1<0【變式3】（2023·高一課時(shí)練習(xí)）用分?jǐn)?shù)指數(shù)冪表示下列各式（式中字母均為正數(shù)）：（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【答案】（1）1；（2）SKIPIF1<0；（3）1.【詳解】（1）原式SKIPIF1<0；（2）原式SKIPIF1<0；（3）原式SKIPIF1<0.題型04條件求值【典例1】（2023秋·河南鄭州·高一鄭州市第七中學(xué)校考期末）已知SKIPIF1<0，下列各式中正確的個(gè)數(shù)是（

）①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；A．1 B．2 C．3 D．4【答案】C【詳解】①SKIPIF1<0,正確；②SKIPIF1<0，正確；③因?yàn)镾KIPIF1<0可知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故錯(cuò)誤；④SKIPIF1<0，正確.故選：C【典例2】（2022秋·江蘇南通·高一江蘇省南通中學(xué)?？计谥校?）求SKIPIF1<0的值；（2）已知SKIPIF1<0，求SKIPIF1<0的值.【答案】（1）SKIPIF1<0；（2）SKIPIF1<0【詳解】（1）SKIPIF1<0；（2）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【典例3】（2022秋·江西萍鄉(xiāng)·高一江西省蓮花中學(xué)?？计谥校┯?jì)算下列各式(1)SKIPIF1<0；(2)已知SKIPIF1<0，求下列各式的值：①SKIPIF1<0；②SKIPIF1<0．【答案】(1)89；(2)①SKIPIF1<0；②SKIPIF1<0.【詳解】（1）原式SKIPIF1<0；（2）①∵SKIPIF1<0，∴SKIPIF1<0，又由SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0；②（法一）SKIPIF1<0SKIPIF1<0，（法二）SKIPIF1<0，而SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，又由SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0.【變式1】（2023·全國(guó)·高三專題練習(xí)）（1）計(jì)算SKIPIF1<0；（2）若SKIPIF1<0，求SKIPIF1<0的值．【答案】（1）-5；（2）14.【詳解】（1）SKIPIF1<00.3﹣1﹣36+33+1SKIPIF1<036+27+1SKIPIF1<05．（2）若SKIPIF1<0，∴xSKIPIF1<02＝6，xSKIPIF1<04，∴x2+x﹣2+2＝16，∴x2+x﹣2＝14．【變式2】（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則SKIPIF1<0．【答案】3【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故答案為：3．【變式3】（2022秋·廣西玉林·高一?？计谥校┮阎猄KIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【詳解】由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0即SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0A夯實(shí)基礎(chǔ)一、單選題1．（2023·全國(guó)·高三專題練習(xí)）化簡(jiǎn)SKIPIF1<0的結(jié)果為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0故選：C.2．（2023·江蘇·高一假期作業(yè)）化簡(jiǎn)SKIPIF1<0的結(jié)果為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0，SKIPIF1<0.故選：D3．（2023·高一課時(shí)練習(xí)）計(jì)算SKIPIF1<0，結(jié)果是（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0.故選：B4．（2023·全國(guó)·高一假期作業(yè)）有下列四個(gè)命題：①正數(shù)的偶次方根是一個(gè)正數(shù)；②正數(shù)的奇次方根是一個(gè)正數(shù)；③負(fù)數(shù)的偶次方根是一個(gè)負(fù)數(shù)；④負(fù)數(shù)的奇次方根是一個(gè)負(fù)數(shù)．其中正確的個(gè)數(shù)是()A．0 B．1C．2 D．3【答案】C【詳解】正數(shù)的偶次方根有兩個(gè)且一正一負(fù)，負(fù)數(shù)的偶次方根不存在；正數(shù)的奇次方根為一個(gè)正數(shù)，負(fù)數(shù)的奇次方根為一個(gè)負(fù)數(shù)；①③錯(cuò)誤，②④正確．故選：C5．（2023·全國(guó)·高一專題練習(xí)）化簡(jiǎn)SKIPIF1<0（a，b為正數(shù)）的結(jié)果是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0.故選：C.6．（2023·全國(guó)·高三專題練習(xí)）2021年5月15日，中國(guó)首次火星探測(cè)任務(wù)天問(wèn)一號(hào)探測(cè)器在火星成功著陸.截至目前，祝融號(hào)火星車在火星上留下1900多米的“中國(guó)腳印”，期待在2050年實(shí)現(xiàn)載人登陸火星.已知所有行星繞太陽(yáng)運(yùn)動(dòng)的軌道都是橢圓，且所有行星軌道的半長(zhǎng)軸的三次方與它的公轉(zhuǎn)周期的二次方的比值都相等.若火星與地球的公轉(zhuǎn)周期之比約為SKIPIF1<0，則地球運(yùn)行軌道的半長(zhǎng)軸與火星運(yùn)行軌道的半長(zhǎng)軸的比值約為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)地球的公轉(zhuǎn)周期為SKIPIF1<0，則火星的公轉(zhuǎn)周期為SKIPIF1<0.設(shè)地球?火星運(yùn)行軌道的半長(zhǎng)軸分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，于是SKIPIF1<0.故選:A.7．（2023·全國(guó)·高一假期作業(yè)）已知SKIPIF1<0，則SKIPIF1<0的值是（

）A．15 B．12 C．16 D．25【答案】A【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又由立方差公式，SKIPIF1<0，故選：A．8．（2023·全國(guó)·高三專題練習(xí)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0.故選：B二、多選題9．（2023·江蘇·高一假期作業(yè)）下列說(shuō)法正確的是（）A．16的4次方根是2B．SKIPIF1<0的運(yùn)算結(jié)果是±2C．當(dāng)n為大于1的奇數(shù)時(shí)，SKIPIF1<0對(duì)任意SKIPIF1<0都有意義D．當(dāng)n為大于1的偶數(shù)時(shí)，SKIPIF1<0只有當(dāng)SKIPIF1<0時(shí)才有意義【答案】CD【詳解】對(duì)于A，由于SKIPIF1<0，所以16的4次方根是SKIPIF1<0，故A不正確．對(duì)于B，SKIPIF1<0，故B不正確．對(duì)于C，由根式的意義知，當(dāng)SKIPIF1<0為大于1的奇數(shù)時(shí)，SKIPIF1<0對(duì)任意SKIPIF1<0都有意義，故C正確．對(duì)于D，由根式的意義知，當(dāng)SKIPIF1<0為大于1的偶數(shù)時(shí)，SKIPIF1<0只有當(dāng)SKIPIF1<0時(shí)才有意義，故D正確．故選：CD．10．（2023·全國(guó)·高一假期作業(yè)）已知SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】AB【詳解】令SKIPIF1<0SKIPIF1<0故選：AB三、填空題11．（2023·高一課時(shí)練習(xí)）求值：SKIPIF1<0．【答案】SKIPIF1<0【詳解】SKIPIF1<0SKIPIF1<0，故答案為:SKIPIF1<0.12．（2023·全國(guó)·高三專題練習(xí)）SKIPIF1<0【答案】SKIPIF1<0【詳解】原式SKIPIF1<0SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0.四、解答題13．（2023·江蘇·高一假期作業(yè)）計(jì)算：(1)SKIPIF1<0；(2)SKIPIF1<0【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0SKIPIF1<0SKIPIF1<0；（2）SKIPIF1<0SKIPIF1<0.14．（2023·江蘇·高一假期作業(yè)）求值：(1)SKIPIF1<0；(2)π0－SKIPIF1<0＋SKIPIF1<0×SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）原式SKIPIF1<0；（2）原式SKIPIF1<0.B能力提升1．（2023·安徽安慶·安徽省桐城中學(xué)校考二模）閱讀下段文字：“已知SKIPIF1<0為無(wú)理數(shù)，若SKIPIF1<0為有理數(shù)，則存在無(wú)理數(shù)SKIPIF1<0，使得SKIPIF1<0為有理數(shù)；若SKIPIF1<0為無(wú)理數(shù)，則取無(wú)理數(shù)SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0為有理數(shù)．”依據(jù)這段文字可以證明的結(jié)論是（

）A．SKIPIF1<0是有理數(shù) B．SKIPIF1<0是無(wú)理數(shù)C．存在無(wú)理數(shù)a，b，使得SKIPIF1<0為有理數(shù) D．對(duì)任意無(wú)理數(shù)a，b，都有SKIPIF1<0為無(wú)理數(shù)【答案】C【詳解】這段文字中，沒有證明SKIPIF1<0是有理數(shù)條件，也沒有證明SKIPIF1<0是無(wú)理數(shù)的條件，AB錯(cuò)誤；這段文字的兩句話中，都說(shuō)明了結(jié)論“存在無(wú)理數(shù)a，b，使得SKIPIF1<0為有理數(shù)”，因此這段文字可以證明此結(jié)論，C正確；這段文字中只提及存在無(wú)理數(shù)a，b，不涉及對(duì)任意無(wú)理數(shù)a，b，都成立的問(wèn)題，D錯(cuò)誤.故選：C2．（2023·江蘇·高一假期作業(yè)）求使等式SKIPIF1<0成立的實(shí)數(shù)a的取值范圍.【答案】[-3，3]【詳解】SKIPIF1<0，要使SKIPIF1<0|成立，需SKIPIF1<0解得a∈[-3，3]．3．（2023·全國(guó)·高一假期作業(yè)）（1）已知SKIPIF1<0，化簡(jiǎn)SKIPIF1<0.（2）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值.【答案】（1）SKIPIF1<0；（2）8【詳解】（1）由SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0.（2）令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.∴SKIPIF1<0.4．（2023·全國(guó)·高三專題練習(xí)）（1）計(jì)算：SKIPIF1<0；（2）已知SKIPIF1<0是方程SKIPIF1<0的兩根，求SKIPIF1<0的值.【答案】（1）16；（2）SKIPIF1<0．【詳解】（1）原式＝SKIPIF1<0SKIPIF1<0；（2）由題意SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，

4.2指數(shù)函數(shù)（4.2.1指數(shù)函數(shù)的概念+4.2.2指數(shù)函數(shù)的圖象和性質(zhì)）課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①了解指數(shù)函數(shù)，掌握指數(shù)函數(shù)的形式及條件，會(huì)根據(jù)底數(shù)區(qū)分兩類函數(shù)。②掌握指數(shù)函數(shù)的圖象與性質(zhì)，能根據(jù)指數(shù)函數(shù)的性質(zhì)進(jìn)行方程、不等式的求解，比較大小，及函數(shù)的單調(diào)區(qū)間的求解、會(huì)求與指數(shù)函數(shù)相關(guān)的函數(shù)的定義域、值域。③能解決與指數(shù)函數(shù)有關(guān)的綜合性問(wèn)題。通過(guò)本節(jié)課的學(xué)習(xí)，要求認(rèn)識(shí)、了解指數(shù)函數(shù)的形式及要求，掌握指數(shù)函數(shù)的圖象與性質(zhì)，并能利用指數(shù)函數(shù)的性質(zhì)進(jìn)行大小的比較、解指數(shù)方程與不等式、會(huì)求復(fù)合函數(shù)的定義域、值域、單調(diào)區(qū)間，能解決與指數(shù)函數(shù)有關(guān)的實(shí)際問(wèn)題及綜合問(wèn)題.知識(shí)點(diǎn)01：指數(shù)函數(shù)的概念1、一般地，函數(shù)SKIPIF1<0叫做指數(shù)函數(shù)，其中指數(shù)SKIPIF1<0是自變量，底數(shù)SKIPIF1<0是一個(gè)大于0且不等于1的常量，定義域是SKIPIF1<0.2、學(xué)習(xí)指數(shù)函數(shù)的定義，注意一下幾點(diǎn)（1）定義域?yàn)椋篠KIPIF1<0（2）規(guī)定SKIPIF1<0是因?yàn)椋孩偃鬝KIPIF1<0，則SKIPIF1<0（恒等于1）沒有研究?jī)r(jià)值；②若SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0（恒等于0），而當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0無(wú)意義；③若SKIPIF1<0，則SKIPIF1<0中SKIPIF1<0為偶數(shù)，SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0無(wú)意義.④只有當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0可以是任意實(shí)數(shù).（3）函數(shù)解析式形式要求：指數(shù)函數(shù)只是一個(gè)新式定義，判斷一個(gè)函數(shù)是指數(shù)函數(shù)的關(guān)鍵有三點(diǎn)：①SKIPIF1<0的系數(shù)必須為1；②底數(shù)為大于0且不等于1的常數(shù)，不能是自變量；③指數(shù)處只有一個(gè)自變量，而不是含自變量的多項(xiàng)式.【即學(xué)即練1】（多選）（2023·全國(guó)·高一假期作業(yè)）下列函數(shù)中，是指數(shù)函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【詳解】由指數(shù)函數(shù)形式為SKIPIF1<0且SKIPIF1<0，顯然A、D不符合，C符合；對(duì)于B，SKIPIF1<0且SKIPIF1<0，故符合.故選：BC知識(shí)點(diǎn)02：指數(shù)函數(shù)的圖象與性質(zhì)1、函數(shù)SKIPIF1<0的圖象和性質(zhì)如下表：底數(shù)SKIPIF1<0SKIPIF1<0圖象性質(zhì)定義域SKIPIF1<0值域SKIPIF1<0定點(diǎn)圖象過(guò)定點(diǎn)SKIPIF1<0單調(diào)性增函數(shù)減函數(shù)函數(shù)值的變化情況當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0對(duì)稱性函數(shù)SKIPIF1<0與SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱2、指數(shù)函數(shù)SKIPIF1<0的底數(shù)SKIPIF1<0對(duì)圖象的影響函數(shù)SKIPIF1<0的圖象如圖所示：觀察圖象，我們有如下結(jié)論：2.1.底數(shù)SKIPIF1<0與1的大小關(guān)系決定了指數(shù)函數(shù)SKIPIF1<0圖象的“升”與“降”.（1）當(dāng)SKIPIF1<0時(shí)，指數(shù)函數(shù)的圖象是“上升”的，且當(dāng)SKIPIF1<0時(shí)，底數(shù)SKIPIF1<0的值越大，函數(shù)的圖象越“陡”，說(shuō)明其函數(shù)值增長(zhǎng)的越快.（2）當(dāng)SKIPIF1<0時(shí)，指數(shù)函數(shù)的圖象是“下降”的，且當(dāng)SKIPIF1<0時(shí)，底數(shù)SKIPIF1<0的值越小，函數(shù)的圖象越“陡”，說(shuō)明其函數(shù)值減小的越快.2.2.底數(shù)SKIPIF1<0的大小決定了圖象相對(duì)位置的高低：不論是SKIPIF1<0還是SKIPIF1<0，底數(shù)越大，在第一象限內(nèi)的函數(shù)圖象越“靠上”.在同一平面直角坐標(biāo)系中，底數(shù)SKIPIF1<0的大小決定了圖象相對(duì)位置的高低；在SKIPIF1<0軸右側(cè)，圖象從上到下相應(yīng)的底數(shù)由大變小，即“底數(shù)大圖象高”；在SKIPIF1<0軸左側(cè)，圖象從上到下相應(yīng)的底數(shù)由小變大，即“底數(shù)大圖象低”；知識(shí)點(diǎn)03：指數(shù)函數(shù)的定義域與值域1、定義域：（1）指數(shù)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0（2）SKIPIF1<0SKIPIF1<0的定義域與函數(shù)SKIPIF1<0的定義域相同（3）SKIPIF1<0的定義域與函數(shù)SKIPIF1<0的定義域不一定相同.2、值域（1）指數(shù)函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0（2）求形如SKIPIF1<0的函數(shù)的值域，先求SKIPIF1<0的值域，然后結(jié)合SKIPIF1<0得性質(zhì)確定SKIPIF1<0的值域（3）求形如SKIPIF1<0的值域，轉(zhuǎn)化為先求SKIPIF1<0的值域，再將SKIPIF1<0的取值范圍代入函數(shù)SKIPIF1<0中.知識(shí)點(diǎn)04：指數(shù)函數(shù)的圖象變換已知函數(shù)SKIPIF1<01、平移變換①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0④SKIPIF1<02、對(duì)稱變換①SKIPIF1<0②SKIPIF1<0③SKIPIF1<03、翻折變換①SKIPIF1<0（去掉SKIPIF1<0軸左側(cè)圖象，保留SKIPIF1<0軸右側(cè)圖象；將SKIPIF1<0軸右側(cè)圖象翻折到SKIPIF1<0軸左側(cè)）②SKIPIF1<0（保留SKIPIF1<0軸上方的圖象，將SKIPIF1<0軸下方的圖象翻折到SKIPIF1<0軸上方）題型01指數(shù)函數(shù)的判定與求值【典例1】（2023秋·四川瀘州·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】A【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A【典例2】（2023·高一課時(shí)練習(xí)）函數(shù)①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0；⑦SKIPIF1<0；⑧SKIPIF1<0中，是指數(shù)函數(shù)的是．【答案】①⑤【詳解】因?yàn)橹笖?shù)函數(shù)為SKIPIF1<0且SKIPIF1<0，故①⑤正確；由冪函數(shù)定義知，SKIPIF1<0是冪函數(shù)，故②不正確；由指數(shù)函數(shù)的定義知，③④⑥⑦均不是指數(shù)函數(shù)；對(duì)于⑧，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不是指數(shù)函數(shù).故答案為：①⑤.【變式1】（2023·高一課時(shí)練習(xí)）設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0＝．【答案】0【詳解】由已知得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.【變式2】（2023·高一課時(shí)練習(xí)）下列函數(shù)中是指數(shù)函數(shù)的是(填序號(hào))．①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0．【答案】③【詳解】①SKIPIF1<0的系數(shù)不是SKIPIF1<0，不是指數(shù)函數(shù)；②SKIPIF1<0的指數(shù)不是自變量SKIPIF1<0，不是指數(shù)函數(shù)；③SKIPIF1<0是指數(shù)函數(shù)；④SKIPIF1<0的底數(shù)是SKIPIF1<0不是常數(shù)，不是指數(shù)函數(shù)；⑤SKIPIF1<0的指數(shù)不是自變量SKIPIF1<0，不是指數(shù)函數(shù)；⑥SKIPIF1<0是冪函數(shù)．故答案為：③題型02根據(jù)函數(shù)是指數(shù)函數(shù)求參數(shù)【典例1】（2023·全國(guó)·高一假期作業(yè)）如果函數(shù)SKIPIF1<0和SKIPIF1<0都是指數(shù)函數(shù)，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．9 D．8【答案】D【詳解】根據(jù)題意可得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0.故選：D【典例2】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0是指數(shù)函數(shù)，則SKIPIF1<0的值為．【答案】SKIPIF1<0【詳解】因?yàn)楹瘮?shù)SKIPIF1<0為指數(shù)函數(shù)，則SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·高一課時(shí)練習(xí)）函數(shù)是指數(shù)函數(shù)SKIPIF1<0，則有SKIPIF1<0.【答案】SKIPIF1<0【詳解】由題意可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0【變式2】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0是指數(shù)函數(shù)，則a的取值范圍是．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0是指數(shù)函數(shù),所以SKIPIF1<0,解得:SKIPIF1<0或SKIPIF1<0即a的取值范圍是SKIPIF1<0.故答案為:SKIPIF1<0題型03指數(shù)型函數(shù)圖象過(guò)定點(diǎn)問(wèn)題【典例1】（多選）（2023·全國(guó)·高一專題練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖像過(guò)定點(diǎn)SKIPIF1<0，則（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0為R上的增函數(shù) D．SKIPIF1<0的解集為SKIPIF1<0【答案】BCD【詳解】由題意可得SKIPIF1<0恒成立，故SKIPIF1<0，A錯(cuò)誤，因?yàn)楦鶕?jù)題意，得SKIPIF1<0，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，故B正確，SKIPIF1<0SKIPIF1<0，所以，SKIPIF1<0為R上的增函數(shù)，C正確；SKIPIF1<0，解得SKIPIF1<0，D正確．故選：BCD【典例2】（2023·全國(guó)·高一假期作業(yè)）函數(shù)SKIPIF1<0SKIPIF1<0且SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0，SKIPIF1<0．【答案】SKIPIF1<0【詳解】令SKIPIF1<0可得SKIPIF1<0，此時(shí)有SKIPIF1<0.由題意可得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0．【變式1】（2023·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0恒過(guò)的定點(diǎn)坐標(biāo)為．【答案】SKIPIF1<0【詳解】解：令SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，所以，函數(shù)SKIPIF1<0恒過(guò)的定點(diǎn)坐標(biāo)為SKIPIF1<0故答案為：SKIPIF1<0【變式2】（2023秋·吉林遼源·高一校聯(lián)考期末）函數(shù)SKIPIF1<0且SKIPIF1<0的圖象恒過(guò)定點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0坐標(biāo)為．【答案】SKIPIF1<0【詳解】令SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，所以定點(diǎn)SKIPIF1<0為SKIPIF1<0，故答案為：SKIPIF1<0題型04指數(shù)函數(shù)圖象的識(shí)別【典例1】（2023春·湖南常德·高一統(tǒng)考期末）指數(shù)函數(shù)SKIPIF1<0的圖象如圖所示，則二次函數(shù)SKIPIF1<0的圖象可能是（

）

A．

B．

C．

D．

【答案】B【詳解】由指數(shù)函數(shù)SKIPIF1<0的圖象可知：SKIPIF1<0.令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，對(duì)應(yīng)只有B選項(xiàng)符合題意.故選：B【典例2】（2023·全國(guó)·高一假期作業(yè)）函數(shù)SKIPIF1<0（SKIPIF1<0）的圖象可能是（

）A．B．C．D．【答案】C【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故A、B均不符合；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0，且函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故C符合，D不符合．故選：C．【變式1】（2023秋·山東臨沂·高一?？计谀┖瘮?shù)SKIPIF1<0的部分圖象大致是（

）A．B．C．D．【答案】B【詳解】SKIPIF1<0定義域?yàn)镽，且SKIPIF1<0，故SKIPIF1<0為偶函數(shù)，關(guān)于y軸對(duì)稱，AC錯(cuò)誤；SKIPIF1<0，SKIPIF1<0，故B正確，D錯(cuò)誤.故選：B.【變式2】（2023·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0(a>0且a≠1)的圖象可能為（

）A．B．C．D．【答案】C【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，顯然當(dāng)SKIPIF1<0時(shí)，函數(shù)單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，函數(shù)單調(diào)遞減，函數(shù)圖象的漸近線為SKIPIF1<0，而SKIPIF1<0，故AB不符合；對(duì)于CD，因?yàn)闈u近線為SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0時(shí)，SKIPIF1<0，故選項(xiàng)C符合，D不符合；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，函數(shù)單調(diào)遞減，函數(shù)圖象的漸近線為SKIPIF1<0，而SKIPIF1<0，故ABD不符合；故選：C題型05畫指數(shù)函數(shù)的圖象【典例1】（2023·全國(guó)·高三對(duì)口高考）利用函數(shù)SKIPIF1<0的圖象，作出下列各函數(shù)的圖象．(1)SKIPIF1<0；(2)SKIPIF1<0(3)SKIPIF1<0；(4)SKIPIF1<0；(5)SKIPIF1<0；(6)SKIPIF1<0．【詳解】（1）把SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱得到SKIPIF1<0的圖象，如圖，

（2）保留SKIPIF1<0圖象在SKIPIF1<0軸右邊部分，去掉SKIPIF1<0軸左側(cè)的，并把SKIPIF1<0軸右側(cè)部分關(guān)于SKIPIF1<0軸對(duì)稱得到SKIPIF1<0的圖象，如圖，

（3）把SKIPIF1<0圖象向下平移一個(gè)單位得到SKIPIF1<0的圖象，如圖，

（4）結(jié)合（3），保留SKIPIF1<0上方部分，然后把SKIPIF1<0下方部分關(guān)于SKIPIF1<0軸翻折得到SKIPIF1<0的圖象，如圖，

（5）把SKIPIF1<0圖象關(guān)于SKIPIF1<0軸對(duì)稱得到SKIPIF1<0的圖象，如圖，

（6）把SKIPIF1<0的圖象向右平移一個(gè)單位得到SKIPIF1<0的圖象，如圖，

【典例2】（2023秋·湖南長(zhǎng)沙·高一校考期末）已知函數(shù)SKIPIF1<0.(1)在平面直角坐標(biāo)系中，畫出函數(shù)SKIPIF1<0的簡(jiǎn)圖，并寫出SKIPIF1<0的單調(diào)區(qū)間和值域；(2)若SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)圖象見解析，SKIPIF1<0的增區(qū)間為SKIPIF1<0，減區(qū)間為SKIPIF1<0，值域?yàn)镾KIPIF1<0.(2)SKIPIF1<0【詳解】（1）函數(shù)SKIPIF1<0的簡(jiǎn)圖如下：由圖可知，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，減區(qū)間為SKIPIF1<0；值域?yàn)镾KIPIF1<0.（2）由SKIPIF1<0，及函數(shù)SKIPIF1<0的單調(diào)性可知，若SKIPIF1<0則實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.【變式1】（2022秋·江西贛州·高一統(tǒng)考期中）著名數(shù)學(xué)家華羅庚曾說(shuō)過(guò)：“數(shù)缺形時(shí)少直觀，形少數(shù)時(shí)難入微；數(shù)形結(jié)合百般好，隔離分家萬(wàn)事休”．在數(shù)學(xué)的學(xué)習(xí)和研究中，常常借助圖象來(lái)研究函數(shù)的性質(zhì)．已知函數(shù)SKIPIF1<0.

(1)在平面直角坐標(biāo)系中作函數(shù)SKIPIF1<0的簡(jiǎn)圖，并根據(jù)圖象寫出該函數(shù)的單調(diào)減區(qū)間；(2)解不等式SKIPIF1<0.【答案】(1)作圖見解析，單調(diào)減區(qū)間為SKIPIF1<0和SKIPIF1<0(2)SKIPIF1<0【詳解】（1）簡(jiǎn)圖如圖所示：由圖可得該函數(shù)的單調(diào)減區(qū)間為SKIPIF1<0和SKIPIF1<0；（2）①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0；綜上：不等式SKIPIF1<0的解集為SKIPIF1<0.【變式2】（2022秋·北京順義·高一校考期中）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的值；(2)畫出函數(shù)SKIPIF1<0的圖象，根據(jù)圖象寫出函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(3)若SKIPIF1<0，求x的取值范圍.【答案】(1)SKIPIF1<0(2)圖象詳見解析，減區(qū)間SKIPIF1<0，增區(qū)間SKIPIF1<0(3)SKIPIF1<0【詳解】（1）SKIPIF1<0.（2）SKIPIF1<0，所以SKIPIF1<0的圖象如下圖所示，由圖可知，SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0（3）SKIPIF1<0，由圖象可知，滿足SKIPIF1<0的SKIPIF1<0的取值范圍是SKIPIF1<0.題型06利用指數(shù)函數(shù)的單調(diào)性比較大小【典例1】（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0在R上單調(diào)遞減，SKIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在R為增函數(shù)，所以SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，綜上，SKIPIF1<0.故選：A.【典例2】（2023·全國(guó)·高三專題練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0