2019_2020学年高中数学第3章指数函数和对数函数3.2.2指数运算的性质课后篇巩固提升（含解析）北师大版.docx

2.2指数运算的性质课后篇巩固提升A组基础巩固1.设a0,将a2a3a2表示成分数指数幂,其结果是()A.a12B.a56C.a76D.a32解析:由题意,a2a3a2=a2-12-13=a76.故选C.答案:C2.若a4+a-4=6,则a2+a-2的值等于()A.6B.6C.2D.22解析:因为(a2+a-2)2=a4+a-4+2a2a-2=a4+a-4+2=6+2=8,且a2+a-20,所以a2+a-2=22.答案:D3.计算1.5-13-760+80.2542+(323)6-2323的结果为()A.110B.89C.97D.121解析:原式=23131+234214+(213312)6-2313=2313+2+2233-2313=2+427=110.答案:A4.化简(-m)2-1m的结果为()A.mB.-m-mC.mmD.m-m解析:由-1m知-1m0,必有m0.又当m0时,m2=|m|=-m,所以(-m)2-1m=m2-mm2=m2-m|m|=m2-m-m=-m-m.答案:B5.下列结论中,正确的个数是()当a0);函数y=(x-2)12-(3x-7)0的定义域是(2,+);若100a=5,10b=2,则2a+b=1.A.0B.1C.2D.3解析:错,(a2)320,而a30;错,当a0,3x-70,得x2且x73,故错;由100a=5得102a=5,又10b=2,102a10b=52=10,102a+b=10.2a+b=1.正确.答案:B6.0.25-12-4-420-116-12=.解析:原式=1416-4-4=-4.答案:-47.若10m=2,10n=3,则1002m-n4的值等于.解析:1002m-n4=(102)2m-n4=102m-n2=10m-n2=10m10n2=10m(10n)12=2312=23=233.答案:2338.83-312-613+333=.解析:原式=83-63-23+3=3.答案:39.导学号85104059计算:(1)(-1.8)0+32-233382-10.01+93;(2)(2a23b12)(-6a12b13)(-3a16b56).解:(1)原式=1+23232782-10.1+36=1+49322-10+33=1+1-10+27=19.(2)原式=2(-6)(-3)a23+12-16b12+13-56=4ab0=4a.10.已知a,b是方程x2-6x+4=0的两个根,且ab0,求a-ba+b的值.解:a,b是方程x2-6x+4=0的两个根,a+b=6,ab=4.ab0,ab0.a-ba+b0.又a-ba+b2=a+b-2aba+b+2ab=6-246+24=210=15,a-ba+b=15=55.B组能力提升1.设a2n=3,a0,则a3n+a-3nan+a-n的值为()A.53B.2C.73D.4114解析:由a2n=3,a0,得an=3,a-n=13,a3n=(3)3=33,a-3n=133.故a3n+a-3nan+a-n=33+1333+13=(33)2+1333+3=2812=73.答案:C2.若a1,b0,且ab+a-b=22,则ab-a-b的值为()A.6B.2或-2C.-2D.2解析:(ab+a-b)2=8,a2b+a-2b=6,(ab-a-b)2=a2b+a-2b-2=4.又aba-b(a1,b0),ab-a-b=2.答案:D3.若102x=25,10y2=5,则10y-x=.解析:由102x=25,得10x=5,10-x=(10x)-1=5-1.又10y2=(10y)12=5,10y=52,故10y-x=10y10-x=525-1=5.答案:54.若a12-a-12=m,则a2+1a=.解析:由a12-a-12=m,两边平方得,a+a-1-2=m2,即a+a-1=m2+2,故a2+1a=a+a-1=m2+2.答案:m2+25.3a92a-33a-73a13=(其中a0).解析:原式=a1392a13-32a12-73a12133=a96-36+76-136=a0=1.答案:16.已知a=-27125,b=20172018,试求a23+33ab+9b23a43-27a13ba133a-33b的值.解:显然a0,所以有原式=a23+3a13b13+(3b13)2a13(a-27b)a13-3b13a13=(a13)3-(3b13)3a23(a-27b)=1a23=a-23=-27125-23=259.7.导学号85104060(拓展探究)已知a0,若对于ar8,rN+,式子(a)8-r1