人教版八年級(jí)數(shù)學(xué)上冊同步練習(xí)：14.1.3積的乘方

1、14.1.3積的乘方一、選擇題1.計(jì)算(-3a)2的結(jié)果是()A.6a2B.-9a2C.9a2D.-6a22.計(jì)算(ab2)3的結(jié)果是()A.3ab2B.ab6C.a3b5D.a3b63.計(jì)算-3a2b34的結(jié)果是()A.81a8b12B.12a6b7 C.-12a6b7D.-81a8b124.下列計(jì)算錯(cuò)誤的是()A.2a2+3a2=5a4B.(3ab3)2=9a2b6C.(x2)3=x6D.a·a2=a35.如果(an·bm·b)3=a9b15,那么()A.m=4,n=3B.m=4,n=4C.m=3,n=4D.m=3,n=36.計(jì)算(12)2021×2

2、2021的結(jié)果是()A.0B.1C.-1D.240427.計(jì)算(-3)2020×(-13)2021的結(jié)果是()A.-1B.1C.-13D.13二、填空題8.計(jì)算:(1)(12xy)3=; (2)-a2(b4)32=. 9.一個(gè)正方體的棱長為4×103 cm,它的體積為k×10a cm3(1k<10,a是正整數(shù)),則k=,a=. 10.下面是佳佳、昊昊兩名同學(xué)計(jì)算(a2·a3)2的過程,他們的解答是否正確?若不正確,請?jiān)诿恳徊胶竺娴臋M線上改正;若正確,請?jiān)诿恳徊胶竺娴臋M線上填上計(jì)算的依據(jù).(1)(a2·a3)2

3、=(a5)2 =a10. 佳佳的解答.(填“正確”或“不正確”) (2)(a2·a3)2=(a2)2·(a3)2  =a4·a5  =a20.  昊昊的解答.(填“正確”或“不正確”) 11.若33x+1×53x+1=152x+4,則x=. 三、解答題12.計(jì)算:(1)(-5ab)3;(2)-(3x2y)2;(3)(-43ab2c3)3;(4)(-xmy3m)2;(5)2a2·a4-3(ab2)2;(6)9m4(n2)3+(-3m2n3

4、)2.13.計(jì)算:(1)-192021×811010;(2)(-0.125)12×(-123)7×(-8)13×(-35)9.14.已知32x+1×22x-32x×22x+1=6x+4,求x的值.15.已知16m=4×22n-2,27n=9×3m+3,求(n-m)2021的值.16.比較218×310與210×315的大小.17.已知P=888888,Q=118880,比較P與Q的大小.答案1.C2.D3.D4.A 5.A6.B7.C8.(1)18x3y3(2)a4b249.6.41010.(1

5、)同底數(shù)冪相乘,底數(shù)不變,指數(shù)相加冪的乘方,底數(shù)不變,指數(shù)相乘正確(2)積的乘方,等于把積的每一個(gè)因式分別乘方,再把所得的冪相乘a4·a6a10不正確11.312.解:(1)(-5ab)3=(-5)3a3b3=-125a3b3.(2)-(3x2y)2=-32x4y2=-9x4y2.(3)-43ab2c33=-433a3b6c9=-6427a3b6c9.(4)(-xmy3m)2=(-1)2x2my6m=x2my6m.(5)原式=2a6-3a2b4.(6)原式=9m4n6+(-3)2·(m2)2·(n3)2=9m4n6+9m4n6=18m4n6.13.解:(1)原式=

6、-192021×92020=-19×92020×-19=-19.(2)原式=(-0.125)×(-8)12×(-8)×(-53)×(-35)7×-352=-8×925=-7225.14.解:32x+1×22x-32x×22x+1=6x+4,3×32x×22x-2×32x×22x=6x+4,3×62x-2×62x=6x+4,62x=6x+4,則2x=x+4,解得x=4.15.解:根據(jù)題意知(24)m=22×22n-2,(33)n=32×3m+3,即24m=22n,33n=3m+5,則4m=2n且3n=m+5,解得m=1,n=2.所以(n-m)2021=(2-1)2021=1.16.解:218×310=28×210×310=28×610,210×315=210×310×35=610×35,且28>35,218×310