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【同步练习】苏科版初一数学下册 第8章《幂的运算》8.2 幂的乘方和积的乘方(专项训练)
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这是一份【同步练习】苏科版初一数学下册 第8章《幂的运算》8.2 幂的乘方和积的乘方(专项训练),共6页。
8.2 幂的乘方和积的乘方(专项训练)1.(2021秋•安岳县期末)若xm=3,xn=2,则x2m+n的值是( )A.11 B.12 C.18 D.362.(2022秋•五华区校级期中)已知ax=﹣2,ay=3,则a3x+2y等于( )A.1 B.72 C.﹣72 D.﹣363.(2022•云安区模拟)已知4n=3,8m=5,则22n+3m=( )A.1 B.2 C.8 D.154.(2022秋•渝北区校级期中)已知3m+2n﹣3=0,则23m×4n的值是( )A.﹣ B. C.﹣8 D.85.已知am=2,an=4,求下列各式的值(1)am+n(2)a3m+2n.6.已知10x=5,10y=6,求103x+2y的值.7.(2022春•宁远县月考)若(xayb)3=x6y15,则a,b的值分别为( )A.2,5 B.3,12 C.5,2 D.12,38.(2022秋•沙坪坝区校级期中)计算(x3y)2的结果是( )A.x5y B.x5y2 C.x6y2 D.x3y29.(2022秋•浦东新区校级期中)计算(﹣)2021•(﹣)2022的结果是( )A. B. C. D.10.计算的结果是( )A. B. C. D.11.计算:a3•a5+(a2)4+(﹣3a4)2.12.x4•x3•x+(x4)2+(﹣2x2)4.答案与解析1.(2021秋•安岳县期末)若xm=3,xn=2,则x2m+n的值是( )A.11 B.12 C.18 D.36【答案】C【解答】解:∵xm=3,xn=2,∴x2m+n=x2m•xn=(xm)2•xn=32×2=18.故选:C.2.(2022秋•五华区校级期中)已知ax=﹣2,ay=3,则a3x+2y等于( )A.1 B.72 C.﹣72 D.﹣36【答案】C【解答】解:当ax=﹣2,ay=3时,a3x+2y=a3x×a2y=(ax)3×(ay)2=(﹣2)3×32=﹣8×9=﹣72.故选:C.3.(2022•云安区模拟)已知4n=3,8m=5,则22n+3m=( )A.1 B.2 C.8 D.15【答案】D【解答】解:当4n=3,8m=5时,22n+3m=22n×23m=(22)n×(23)m=4n×8m=3×5=15.故选:D.4.(2022秋•渝北区校级期中)已知3m+2n﹣3=0,则23m×4n的值是( )A.﹣ B. C.﹣8 D.8【答案】D【解答】解:∵3m+2n﹣3=0,∴3m+2n=3,∴23m×4n=23m×22n=23m+2n=23=8.故选:D.5.已知am=2,an=4,求下列各式的值(1)am+n(2)a3m+2n.【解答】解:(1)∵am=2,an=4,∴am+n=am×an=2×4=8;(2)∵am=2,an=4,∴a3m+2n=(am)3×(an)2=8×16=128.6.已知10x=5,10y=6,求103x+2y的值.【解答】解:∵10x=5,10y=6,∴103x+2y=103x•102y=(10x)3•(10y)2=53×62=4500.7.(2022春•宁远县月考)若(xayb)3=x6y15,则a,b的值分别为( )A.2,5 B.3,12 C.5,2 D.12,3【答案】A【解答】解:∵(xayb)3=x6y15,∴x3ay3b=x6y15,∴3a=6,3b=15,∴a=2,b=5,故选:A.8.(2022秋•沙坪坝区校级期中)计算(x3y)2的结果是( )A.x5y B.x5y2 C.x6y2 D.x3y2【答案】C【解答】解:(x3y)2=x6y2.故选:C.9.(2022秋•浦东新区校级期中)计算(﹣)2021•(﹣)2022的结果是( )A. B. C. D.【答案】B【解答】解:原式=[(﹣)×(﹣)]2021×(﹣)=12021×(﹣)=﹣,故选:B.10.计算的结果是( )A. B. C. D.【答案】C【解答】解:===,故选:C.11.计算:a3•a5+(a2)4+(﹣3a4)2.【解答】解:a3•a5+(a2)4+(﹣3a4)2=a8+a8+9a8=11a812.x4•x3•x+(x4)2+(﹣2x2)4.【解答】解:x4•x3•x+(x4)2+(﹣2x2)4=x8+x8+16x8=18x8.
8.2 幂的乘方和积的乘方(专项训练)1.(2021秋•安岳县期末)若xm=3,xn=2,则x2m+n的值是( )A.11 B.12 C.18 D.362.(2022秋•五华区校级期中)已知ax=﹣2,ay=3,则a3x+2y等于( )A.1 B.72 C.﹣72 D.﹣363.(2022•云安区模拟)已知4n=3,8m=5,则22n+3m=( )A.1 B.2 C.8 D.154.(2022秋•渝北区校级期中)已知3m+2n﹣3=0,则23m×4n的值是( )A.﹣ B. C.﹣8 D.85.已知am=2,an=4,求下列各式的值(1)am+n(2)a3m+2n.6.已知10x=5,10y=6,求103x+2y的值.7.(2022春•宁远县月考)若(xayb)3=x6y15,则a,b的值分别为( )A.2,5 B.3,12 C.5,2 D.12,38.(2022秋•沙坪坝区校级期中)计算(x3y)2的结果是( )A.x5y B.x5y2 C.x6y2 D.x3y29.(2022秋•浦东新区校级期中)计算(﹣)2021•(﹣)2022的结果是( )A. B. C. D.10.计算的结果是( )A. B. C. D.11.计算:a3•a5+(a2)4+(﹣3a4)2.12.x4•x3•x+(x4)2+(﹣2x2)4.答案与解析1.(2021秋•安岳县期末)若xm=3,xn=2,则x2m+n的值是( )A.11 B.12 C.18 D.36【答案】C【解答】解:∵xm=3,xn=2,∴x2m+n=x2m•xn=(xm)2•xn=32×2=18.故选:C.2.(2022秋•五华区校级期中)已知ax=﹣2,ay=3,则a3x+2y等于( )A.1 B.72 C.﹣72 D.﹣36【答案】C【解答】解:当ax=﹣2,ay=3时,a3x+2y=a3x×a2y=(ax)3×(ay)2=(﹣2)3×32=﹣8×9=﹣72.故选:C.3.(2022•云安区模拟)已知4n=3,8m=5,则22n+3m=( )A.1 B.2 C.8 D.15【答案】D【解答】解:当4n=3,8m=5时,22n+3m=22n×23m=(22)n×(23)m=4n×8m=3×5=15.故选:D.4.(2022秋•渝北区校级期中)已知3m+2n﹣3=0,则23m×4n的值是( )A.﹣ B. C.﹣8 D.8【答案】D【解答】解:∵3m+2n﹣3=0,∴3m+2n=3,∴23m×4n=23m×22n=23m+2n=23=8.故选:D.5.已知am=2,an=4,求下列各式的值(1)am+n(2)a3m+2n.【解答】解:(1)∵am=2,an=4,∴am+n=am×an=2×4=8;(2)∵am=2,an=4,∴a3m+2n=(am)3×(an)2=8×16=128.6.已知10x=5,10y=6,求103x+2y的值.【解答】解:∵10x=5,10y=6,∴103x+2y=103x•102y=(10x)3•(10y)2=53×62=4500.7.(2022春•宁远县月考)若(xayb)3=x6y15,则a,b的值分别为( )A.2,5 B.3,12 C.5,2 D.12,3【答案】A【解答】解:∵(xayb)3=x6y15,∴x3ay3b=x6y15,∴3a=6,3b=15,∴a=2,b=5,故选:A.8.(2022秋•沙坪坝区校级期中)计算(x3y)2的结果是( )A.x5y B.x5y2 C.x6y2 D.x3y2【答案】C【解答】解:(x3y)2=x6y2.故选:C.9.(2022秋•浦东新区校级期中)计算(﹣)2021•(﹣)2022的结果是( )A. B. C. D.【答案】B【解答】解:原式=[(﹣)×(﹣)]2021×(﹣)=12021×(﹣)=﹣,故选:B.10.计算的结果是( )A. B. C. D.【答案】C【解答】解:===,故选:C.11.计算:a3•a5+(a2)4+(﹣3a4)2.【解答】解:a3•a5+(a2)4+(﹣3a4)2=a8+a8+9a8=11a812.x4•x3•x+(x4)2+(﹣2x2)4.【解答】解:x4•x3•x+(x4)2+(﹣2x2)4=x8+x8+16x8=18x8.
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