2020-2021学年北师大版七年级数学下册第一章
1.7.1单项式除以单项式
同步练习题
A组(基础题)
一、填空题
1.计算:
(1)2x2y3÷(-3xy)=___________;
(2)10x2y3÷2x2y=___________;
(3)3x4y5÷(-xy2)=___________;
2.(1)化简:(2a2b)3÷(-2ab)=___________;
(2)计算:-21x2y4·6x3÷(-3x2y)2=___________;
3.(1)如果3a3b2÷A=ab,那么A=___________;
(2)计算:20x3y5z÷(-10x2y)-2xy·y3z=___________;
4.(1)已知4x2y3÷(-xy)2=16,则y的值为___________;
(2)已知(-a3)4÷a12·(-2a3)=-54,则a的值为___________;
二、选择题
5.计算(-2a3)2÷a2的结果是(
)
A.-2a3
B.-2a4
C.4a3
D.4a4
6.已知8a3bm÷8anb2=b2,那么m,n的值分别为(
)
A.m=4,n=3
B.m=4,n=1
C.m=1,n=3
D.m=2,n=3
7.计算:20x14y4÷(-2x3y)2÷(5xy2)=(
)
A.-x6
B.y4
C.-x7
D.x7
8.下列算式中,不正确的是(
)
A.(-12a5b)÷(-3ab)=4a4
B.9xmyn-1÷(xm-2yn-3)=27x2y2
C.a2b3÷(ab)=ab2
D.x(x-y)2÷(y-x)=-x(x-y)
三、解答题
9.计算:
(1)(-2xy2)2÷3xy;
(2)3m2·8m3n2÷6m7;
(3)(-ab2)3·(-9a3b)÷(-3a3b5);
(4)(-6x4y7)÷(-2xy2)÷(-3x2y4).
10.计算:
(1)24x2y÷(-6xy)+8x5÷2x4;
(2)(-5r3)2÷5r4+(1-2r)(1+2r);
(3)(-2a2b3)÷(-ab2)-b(3a-1);
(4)(2x3y)2·(-2xy)+(-2x3y)3÷2x2.
B组(中档题)
一、填空题
11.(1)计算:(2a+b)4÷(2a+b)2-4a(a+b)=___________;
(2)若一个长方形的面积为a2bc,长为ac,则它的宽为___________;
12.(1)计算:(x-y)5÷(y-x)2=___________;
(2)已知|m-3|+(n-2)2=0,化简:6am+5bm÷(-2abn)=___________;
13.计算:
(1)(a2-b2)2÷(a-b)2=___________;
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy)=___________;
二、解答题
14.(1)已知a4=5,求(-a5)4÷a12·(-2a4)的值.
(2)已知(2amb3)2÷(-a2bn)=ka6b4,求m-n+k-1的值.
C组(综合题)
15.已知(-xyz)2·M=x2n+2yn+3z4÷5x2n-1yn+1z,且自然数x,z满足2x·3z-1=72,求M的值.
参考答案
2020-2021学年北师大版七年级数学下册第一章
1.7.1单项式除以单项式
同步练习题
A组(基础题)
一、填空题
1.计算:
(1)2x2y3÷(-3xy)=-xy2;
(2)10x2y3÷2x2y=5y2;
(3)3x4y5÷(-xy2)=-x3y3.
2.(1)化简:(2a2b)3÷(-2ab)=-4a5b2.
(2)计算:-21x2y4·6x3÷(-3x2y)2=-14xy2.
3.(1)如果3a3b2÷A=ab,那么A=9a2b.
(2)计算:20x3y5z÷(-10x2y)-2xy·y3z=-4xy4z.
4.(1)已知4x2y3÷(-xy)2=16,则y的值为1.
(2)已知(-a3)4÷a12·(-2a3)=-54,则a的值为3.
二、选择题
5.计算(-2a3)2÷a2的结果是(D)
A.-2a3
B.-2a4
C.4a3
D.4a4
6.已知8a3bm÷8anb2=b2,那么m,n的值分别为(A)
A.m=4,n=3
B.m=4,n=1
C.m=1,n=3
D.m=2,n=3
7.计算:20x14y4÷(-2x3y)2÷(5xy2)=(D)
A.-x6
B.y4
C.-x7
D.x7
8.下列算式中,不正确的是(C)
A.(-12a5b)÷(-3ab)=4a4
B.9xmyn-1÷(xm-2yn-3)=27x2y2
C.a2b3÷(ab)=ab2
D.x(x-y)2÷(y-x)=-x(x-y)
三、解答题
9.计算:
(1)(-2xy2)2÷3xy;
解:原式=4x2y4÷3xy=xy3.
(2)3m2·8m3n2÷6m7;
解:原式=24m5n2÷6m7=4m-2n2.
(3)(-ab2)3·(-9a3b)÷(-3a3b5);
解:原式=(-a3b6)·(-9a3b)÷(-3a3b5)
=9a6b7÷(-3a3b5)
=-3a3b2.
(4)(-6x4y7)÷(-2xy2)÷(-3x2y4).
解:原式=-xy.
10.计算:
(1)24x2y÷(-6xy)+8x5÷2x4;
解:原式=0.
(2)(-5r3)2÷5r4+(1-2r)(1+2r);
解:原式=r2+1.
(3)(-2a2b3)÷(-ab2)-b(3a-1);
解:原式=b.
(4)(2x3y)2·(-2xy)+(-2x3y)3÷2x2.
解:原式=4x6y2·(-2xy)+(-8x9y3)÷2x2
=-8x7y3-4x7y3
=-12x7y3.
B组(中档题)
一、填空题
11.(1)计算:(2a+b)4÷(2a+b)2-4a(a+b)=b2.
(2)若一个长方形的面积为a2bc,长为ac,则它的宽为5ab.
12.(1)计算:(x-y)5÷(y-x)2=(x-y)3.
(2)已知|m-3|+(n-2)2=0,化简:6am+5bm÷(-2abn)=-3a7b.
13.计算:
(1)(a2-b2)2÷(a-b)2=a2+2ab+b2;
(2)[(xy+2)(xy-2)-2(x2y2-2)]÷(xy)=-xy.
二、解答题
14.(1)已知a4=5,求(-a5)4÷a12·(-2a4)的值.
解:(-a5)4÷a12·(-2a4)=a20÷a12·(-2a4)
=-2a12.
∵a4=5,∴原式=-2×(a4)3=-2×53=-250.
(2)已知(2amb3)2÷(-a2bn)=ka6b4,求m-n+k-1的值.
解:左边=(2amb3)2÷(-a2bn)=-8a2m-2b6-n,
∵左边=右边,即-8a2m-2b6-n=ka6b4,
∴k=-8,2m-2=6,6-n=4,即k=-8,m=4,n=2.
∴m-n+k-1=4-2+(-8)-1=-=-.
C组(综合题)
15.已知(-xyz)2·M=x2n+2yn+3z4÷5x2n-1yn+1z,且自然数x,z满足2x·3z-1=72,求M的值.
解:由题意,得M=x2n+2yn+3z4÷5x2n-1yn+1z÷(-xyz)2=x3y2z3÷x2y2z2=xz.
∵自然数x,z满足2x·3z-1=72=23×32,
∴x=3,z-1=2.
∴x=3,z=3.
因此,M=×3×3=.