41指數(shù)-2020-2021學(xué)年高一數(shù)學(xué)練習(xí)和分類專題教案（人教A版2019）

第四章指數(shù)函數(shù)與對數(shù)函數(shù)課時4.1指數(shù)1.理解n次方根及根式的概念,掌握根式的性質(zhì).2.掌握指數(shù)冪的運(yùn)算性質(zhì).3.體會數(shù)學(xué)抽象的過程,加強(qiáng)邏輯推理、數(shù)學(xué)運(yùn)算素養(yǎng)的培養(yǎng).基礎(chǔ)過關(guān)練題組一根式的概念及其性質(zhì)1.已知(a-b)A.a>b B.a≥b C.a<b D.a≤b2.已知xy≠0且4x2yA.xy<0 B.xy>0C.x>0,y>0 D.x<0,y>03.在①4(-4)2n,②4(-4)A.①② B.①③C.②③④ D.①②④4.已知a<b<0,n>1,n∈N*,化簡n(a-題組二分?jǐn)?shù)指數(shù)冪與根式的運(yùn)算5.計(jì)算:12-1+82A.6B.7 C.8 D.36.設(shè)a>0,將a2aA.a12 B.a56 C.a7.計(jì)算:(27)23×9A.3 B.13 C.3D.8.化簡[3(-5A.5 B.5 C.5 D.59.若x3=3xA.927 B.327 C.10.化簡:(a2·5a3)÷(a·10a題組三指數(shù)冪的條件求值問題11.若102x=25,則10x等于()A.15 B.25 C.45 12.若a>0,且ax=3,ay=5,則a2x+13.設(shè)α,β是方程5x2+10x+1=0的兩個實(shí)數(shù)根,則2α·2β=,(2α)β=.

14.若x2+2x+1+y2+6y+9=0,則(x2019)15.已知a,b是方程x26x+4=0的兩個實(shí)數(shù)根,且a>b>0,求a-能力提升練題組一根式的概念及其性質(zhì)1.若a<14,則化簡4(4A.4a-1 C.4a-1 2.(多選)下列運(yùn)算結(jié)果中,一定正確的是()A.a3·a4=a7 B.(-a2C.8a8=a D.53.化簡:(a-1)2+(1題組二分?jǐn)?shù)指數(shù)冪與根式的運(yùn)算4.下列各式中正確的是()A.nm7=n7m17 C.4x3+y3=(x+y)35.化簡:5x-23y6.計(jì)算:(32×3)6+(22)434×1649-7.計(jì)算:(1)(0.008(2)(2a13b12)(6a1題組三指數(shù)冪的條件求值問題8.若0<a<1,b>0,且ab+ab=22,則abab等于()A.6 B.2或2 C.2 D.29.已知a+a1=3,下列各式中正確的個數(shù)是()①a2+a2=7;②a3+a3=18;③a12+a-12=±5;④aaA.1 B.2 C.3 D.410.設(shè)a∈R,且a12a-12=2,則a11.計(jì)算:21412(9.6)03312.已知x+x1=4,其中0<x<1,求x2

答案全解全析基礎(chǔ)過關(guān)練1.B因?yàn)?a-b)2.A因?yàn)閤y≠0且4x23.B由(4)2n>0知①有意義;由(4)2n+1<0知②無意義;由a4≥0知③有意義;當(dāng)a<0時,a5<0,此時④無意義.故選B.4.解析當(dāng)n是奇數(shù)時,原式=(ab)+(a+b)=2a;當(dāng)n是偶數(shù)時,因?yàn)閍<b<0,所以ab<0,a+b<0,所以原式=|ab|+|a+b|=(ba)+(ab)=2a.所以n(a=2a5.B12-1+8故選B.6.Ca2a·3a2=a2a·a23=7.D(27)23×9-32=3(-27)2×(32)-8.B原式=(52)13×34=59.A由x3=3x,得x34=3x?x10.答案a解析(a2·5a3)÷(a·=a2·a35÷(a12·a911.A由102x=25可得10x=5,所以10x=1512.答案95解析因?yàn)閍>0,所以a2x+y2=(ax)2·(13.答案14;解析利用一元二次方程根與系數(shù)的關(guān)系,得α+β=2,αβ=15.則2α·2β=2α+β=22=14,(2α)β=2αβ=14.答案1解析因?yàn)閤2+2x+1+所以(x+1)2+所以(x2019)y=[(1)2019]3=(1)3=1.15.解析因?yàn)閍,b是方程x26x+4=0的兩個實(shí)數(shù)根,所以a+b=6,ab=4,所以a-ba+b2=a+b-2能力提升練1.B∵a<14,∴4a∴4(4a2.ADa3·a4=a3+4=a7,故A正確;(a2)3=a6,故B不正確;8a8=|a|,故C不正確;53.解析依題意得a1≥0,即a≥1.∴原式=a1+|1a|+(a)=a11+aa=a2.4.D對于選項(xiàng)A,nm7=n7m7,故A錯誤;對于選項(xiàng)B,12(-3)4=1234=33,故B錯誤;對于選項(xiàng)C,4x3+y3=(x35.答案6x解析5x-23y12-12x-6.解析原式=(213×312)6+(232=22×33+234×4=108+273=100.7.解析(1)(0.0081)-=(34×10-4)-1=31×1013×1(2)(2a13b12)(6a12b12)÷(3a18.C由ab+ab=22,得(ab+ab)2=a2b+a2b+2=8,因此a2b+a2b=6,所以(abab)2=a2b+a2b2=4,由題意得0<ab<1,ab>1,故abab<0,所以abab=2.故選C.9.C由a+a1=3,得(a+a1)2=a2+a2+2=9,所以a2+a2=7,故①正確;因?yàn)閍3+a3=(a+a1)(a2a·a1+a2)=3×(71)=18,所以②正確;因?yàn)?a12+a-12所以a12+a-由aa+1aa又a>0,∴aa+1aa=2因此選C.10.答案42解析∵a12∴a>1且(a12a-12∴a+a1=6,∴(a+a1)2=a2+a2+2=36,∴a2+a2=34,因此(aa1)2=a2+a22=342=32,又a>1,∴aa1>0,∴aa1=32=42.11.解析原式=94121