七年级数学下册1.3.1同底数幂的除法同步练习新版北师大版.docx

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1、 1.3.1 同底数幂的除法一、选择题:1.下列运算结果正确的是( )2x -x =x x(x) =x(-x)(-x) =x (0.1) 10 =103323521363-2-1A.B.C.D.11(- )(- )0 , 则( )2.若 a=-0.3 ,b=-3 ,c=-2-2 ,d=233A.abcdB.badcC.adcbD.cadQB.P=QC.PQD.无 法确定5.已知 a0,下列等式不 正确的是( )1B.(a + )=11( ) =1A.(-7a) =10C.(a-1) =1 D.00202a3 = 5,3 = 43,则 2m-n6.若等于( )mn25A.B.6C.21D.204

2、一、填空题:(-x) (-x)x x x x4 =_.7.计算8.水的质量 0.000204kg,用科学记数法表示为_.(x - 2)0 有意义,则 x_.52 =_, 10239.若(3- ) + (-0.2)-2 =_.10.11.p 0(m - n) (m - n) (m - n)23 24 =_.10 1012.若 5x-3y-2=0,则=_.2 =_.5x3 y= 3,a = 913.如果am,则a3m- nn9 27 3= 81,那么 m=_.14.如果m+3m+14m+7 91016( ) ( ) ( ) = 215.若整数 x、y、z满足,则 x=_,y=_,z=_.xyx89

3、15721(5a -b) (5a -b) = 2416.2,则 m、n的关系(m,n为自然数)是_.mn8三、解答题:17.计算:21( ) + (-1) + ( ) -3(-27) (-9) (-3)-7 ;(1)(3)(4)033;(2)-1520-3365321( ) ( ) + (- ) ( ) - ( -3) + 33-32-30-1 .56233(x + y) (-x - y)2n+1 (n是正整数).2n418.若(3x+2y-10)0无意义,且 2x+y=5,求 x、y的值.(6分)2 - (4 +16 ).(6分)19.化简: 412n+nn3 = 5,3 =1099,求(1

4、) ; (2) 2 .(6分)m-n20.已知 2mnm-n+ x = mx + x221.已知 x,求的值.(6分)-1-2(x -1) =122.已知2 ,求整数 x.(6分)x+ 参考答案1.B 2.B 3.C 4.B 5.C 6.A 7.-x ,x 8.2.0410 kg 9.2-43110.26 11.(m-n) 12.100 13. 14.2 15.3,2,2 16.2m=n63-(x + y)6n-117.(1)9 (2)9 (3)1 (4)18.x=0,y=519.019 = (3 ) = 3= 3 3 = 5100 =20.(1)22222n.m-nm-nm- nm2019(

5、2) 2= (3 )= (3 ) (3 ) = 5 10 =2 .m-n2 2m-n2m2n224+ x = (x + x ) - 2 = m - 221.x2-2-1 2222.当 x+2=0时,x+10,x=-2当 x-1=1时,x=2当 x-1=-1时,x+2为偶数,这时 x=0整数 x为-2,0,2.91016( ) ( ) ( ) = 215.若整数 x、y、z满足,则 x=_,y=_,z=_.xyx8915721(5a -b) (5a -b) = 2416.2,则 m、n的关系(m,n为自然数)是_.mn8三、解答题:17.计算:21( ) + (-1) + ( ) -3(-27)

6、 (-9) (-3)-7 ;(1)(3)(4)033;(2)-1520-3365321( ) ( ) + (- ) ( ) - ( -3) + 33-32-30-1 .56233(x + y) (-x - y)2n+1 (n是正整数).2n418.若(3x+2y-10)0无意义,且 2x+y=5,求 x、y的值.(6分)2 - (4 +16 ).(6分)19.化简: 412n+nn3 = 5,3 =1099,求(1) ; (2) 2 .(6分)m-n20.已知 2mnm-n+ x = mx + x221.已知 x,求的值.(6分)-1-2(x -1) =122.已知2 ,求整数 x.(6分)x

7、+ 参考答案1.B 2.B 3.C 4.B 5.C 6.A 7.-x ,x 8.2.0410 kg 9.2-43110.26 11.(m-n) 12.100 13. 14.2 15.3,2,2 16.2m=n63-(x + y)6n-117.(1)9 (2)9 (3)1 (4)18.x=0,y=519.019 = (3 ) = 3= 3 3 = 5100 =20.(1)22222n.m-nm-nm- nm2019(2) 2= (3 )= (3 ) (3 ) = 5 10 =2 .m-n2 2m-n2m2n224+ x = (x + x ) - 2 = m - 221.x2-2-1 2222.当

8、 x+2=0时,x+10,x=-2当 x-1=1时,x=2当 x-1=-1时,x+2为偶数,这时 x=0整数 x为-2,0,2.91016( ) ( ) ( ) = 215.若整数 x、y、z满足,则 x=_,y=_,z=_.xyx8915721(5a -b) (5a -b) = 2416.2,则 m、n的关系(m,n为自然数)是_.mn8三、解答题:17.计算:21( ) + (-1) + ( ) -3(-27) (-9) (-3)-7 ;(1)(3)(4)033;(2)-1520-3365321( ) ( ) + (- ) ( ) - ( -3) + 33-32-30-1 .56233(x

9、 + y) (-x - y)2n+1 (n是正整数).2n418.若(3x+2y-10)0无意义,且 2x+y=5,求 x、y的值.(6分)2 - (4 +16 ).(6分)19.化简: 412n+nn3 = 5,3 =1099,求(1) ; (2) 2 .(6分)m-n20.已知 2mnm-n+ x = mx + x221.已知 x,求的值.(6分)-1-2(x -1) =122.已知2 ,求整数 x.(6分)x+ 参考答案1.B 2.B 3.C 4.B 5.C 6.A 7.-x ,x 8.2.0410 kg 9.2-43110.26 11.(m-n) 12.100 13. 14.2 15.3,2,2 16.2m=n63-(x + y)6n-117.(1)9 (2)9 (3)1 (4)18.x=0,y=519.019 = (3 ) = 3= 3 3 = 5100 =20.(1)22222n.m-nm-nm- nm2019(2) 2= (3 )= (3 ) (3 ) = 5 10 =2 .m-n2 2m-n2m2n224+ x = (x + x ) - 2 = m - 221.x2-2-1 2222.当 x+2=0时,x+10,x=-2当 x-1=1时,x=2当 x-1=-1时,x+2为偶数,这时 x=0整数 x为-2,0,2.