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1、北师大版七年级数学下册第一章整式的乘除第一节同底数幂的乘法专项练习题(含答案)1. x5 (xm )n 的计算结果是: ( )A. xm+n+5B.x5mnC.x5+mnD.x3(m+n)2. 下列运算正确的是 ()A 2 3= 6B(y2) 3=y6C(m2n) 3=m5n3D2x2+5x2=3x2 3若3m=2,3n=5,则3m+n 的值是()A7B90C10Da2b 4计算(-0.25)2015 42015 的结果是()5. 下列计算中,正确的是()A. a2 + a2 = a4(a2 )3 = a8B.a2 a3 = a66. 下列运算正确的是()A 0.25B -0.25C1D1Ca
2、 2a-2 = a4DA6a5a=1B(a2)3=a5C3a2+2a3=5a5Da6a2=a8 7计算 a2a3 的正确结果是()A. a5Ba6Ca8Da98计算9用科学计数法表示 0.000000001=。10计算 2x3x2 的结果是11已知 3n27=320,则 n=12若 3x=4,3y=7,则 3x+y 的值为13 2( )2=14 x2 x6 ; (- y2 ) y5 ;15比较218 310 与210 315 的大小.16. 阅读计算:阅读下列各式:,回答下列三个问题:(1)验证:(50.2)10=;5100.210=(2)通过上述验证,归纳得出:=;=(3)请应用上述性质计算
3、:17. 解方程组或计算:(1) x -1 + y +1 = 1(2) 23x + y = 418. 已知,求 x 的值.19. 计算: 100 103 102 .答案1C 解: x5 (xm )n = x5 xmn = x5+mn .故选 C. 2D解:根据同底数幂相乘,底数不变,指数相加,可知 2 3= 5,故不正确; 根据幂的乘方,可知(y2) 3=-y6,故不正确;根据积的乘方,等于各个因式分别乘方,可知(m2n) 3=m6n3,故不正确;根据合并同类项法则,可知2x2+5x2=3x2,故正确.故选:D3C 解:3m=2,3n=5, 3m+n = 3m 3n = 2 5 =10.故选
4、C.4D 解:原式=(-0.25)42015=-1故选 D5C解:A、合并同类项系数相加字母及指数不变,a2+a2=2a2,故 A 不正确;B、同底数幂的乘法底数不变指数相加,a2a3=a5,故 B 不正确;C、同底数幂的除法底数不变指数相减,故 C 正确;D、同底数幂的除法底数不变指数相减,故 D 错误;故选 C6D解:A、应为 6a-5a=a,故本选项错误; B、应为(a2)3=a23=a6,故本选项错误;C、3a2 与 2a3 不是同类项,不能合并,故本选项错误; D、a6a2=a8,正确故选 D7A 解:故选 A.8解:原式=.9110-9 解:0.000000001 用科学计数法表示
5、为 110-9,故答案为:110-910 2x5解:根据单项式乘以单项式,结合同底数幂相乘,底数不变,指数相加,可知2x3x2=2x3+2=2x5.故答案为:2x51117 解:3n27=320,3n33=320,3n+3=320,n+3=20,解得 n=17.故答案为:17.1228 解:3x=4,3y=7,3x+y=3x3y=47=2813a4解:根据乘方的意义和同底数幂相乘,直接可得 2( )2=a4.故答案为: 4.14 x4- y3解: x2 x4 x6 ; (- y2 ) (- y3 ) y5 ;15 218 310 210 315解 :Q 218 310 = 210 310 28
6、 ,210 315 = 210 310 35 ,28 35 , 218 310 210 315 .16(1)1,1(2),(3)4,-0.125解 :(1)(50.2)10=110=1; 5100.210=(50.2)10=110=1.(2)=anbn;= anbncn(3)=4=(-0.125)=(-0.125)1=-0.125.x = -117(1) -4x6 (2) y = 5解:(1)原式=-8x6+4x6=-4x6;(2)由 x+y=4 得,x=4-y,代入第一个方程得 y=5, 把 y=5 代入 x=4-y 得,x=-1,x = -1方程组的解为:18y = 5解:19107解:原
7、式=102 103 102 = 102+3+2 = 107 .“”“”At the end, Xiao Bian gives you a passage. Minand once said, people who learn to learn are very happy people. In every wonderful life, learning is an eternal theme. As a professional clerical and teaching position, I understand the importance of continuous learning
8、, life is diligent, nothing can be gained, only continuous learning can achieve better self. Only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. This document is also edited by my studio professionals, there may be errors in the document, if there are errors, please correct, thank you!