专题3.2 单项式的乘法(专项训练)
1.(东方期末)计算:3a2b•(﹣2ab2)= .
【答案】﹣6a3b3
【解答】解:3a2b•(﹣2ab2)
=﹣6a3b3.
故答案为:﹣6a3b3.
2.(东丽区期末)计算3x2•5x3的结果等于 .
【答案】15x5
【解答】解:3x2•5x3=15x2+3=15x5,
故答案为:15x5.
3.(花都区期末)计算 a2•(﹣6ab)的结果是 .
【答案】﹣2a3b
【解答】解: a2•(﹣6ab)
= ×(﹣6)a2+1b
=﹣2a3b.
故答案为:﹣2a3b.
4.(宁国市校级月考)如果单项式﹣3m6﹣2bn2a+b与m1n18分是同类项,那么这两个单项式的积是( )
A.﹣3m2n36 B.﹣3m6n16 C.﹣3m3n8 D.﹣9m6n16
【答案】A
【解答】解:∵单项式﹣3m6﹣2bn2a+b与m1n18是同类项,
单项式﹣3m6﹣2bn2a+b与m1n18分别是单项式﹣3mn18与mn18,
则这两个单项式的积是﹣3mn18•mn18=﹣3m2n36.
故选:A.
5.(江油市期中)计算:4x2y(﹣xy2)3.
【解答】解:原式=4x2y•(﹣x3y6)
=﹣4x5y7.
6.(平潭县校级期中)计算:
(1)5a2•(﹣3a3)2;
(2)( )2021•(﹣2)2022.
【解答】解:(1)5a2•(﹣3a3)2
=5a2•9a6
=45a8;
(2)( )2021•(﹣2)2022
=( )2021•22021•2
=( ×2)2021×2
=1×2
=2.
7.(闵行区期中)计算:(﹣2x2)3+3x•x2•x3+(﹣x3)2.
【解答】解:(﹣2x2)3+3x•x2•x3+(﹣x3)2
=﹣8x6+3x6+x6
=﹣4x6.
8.(通州区校级月考)计算:
(1)(﹣3a2b)2•2ab2;
(2)(m﹣n)•(n﹣m)3•(n﹣m)4;
(3)( )2015×(﹣1.25)2016;
(4)(﹣2a2b)3+4(﹣ab)2(2a4b).
【解答】解:(1)原式=9a4b2•2ab2
=18a5b4;
(2)原式=﹣(m﹣n)4•(m﹣n)4
=﹣(m﹣n)8;
(3)原式=(﹣ )
=(﹣1)
=﹣1×
= ;
(4)原式=﹣8a6b3+4a2b2•(2a4b)
=﹣8a6b3+8a6b3
=0.
9.(海门市校级月考)计算:
(1)2a•6a2;
(2)(﹣4xy3)(﹣2x2);
(3)(3×102)×(5×105).
【解答】解:(1)原式=(2×6)a1+2
=12a3;
(2)原式=[﹣4×(﹣2)]x1+2y3
=8x3y3;
(3)原式=1.5×108.
10.(巨野县期中)计算
(1)(﹣3xy)2(﹣ x2y)3•(﹣ yz2)2;
(2)﹣3xy[6xy﹣3(xy﹣ x2y)].
【解答】解:(1)(﹣3xy)2(﹣ x2y)3•(﹣ yz2)2
=9x2y2•(﹣ x6y3)• y2z4
=﹣ x8y5• y2z4
=﹣ x8y7z4;
(2)﹣3xy[6xy﹣3(xy﹣ x2y)]
=﹣3xy[6xy﹣3xy+x2y)]
=﹣18x2y2+9x2y2﹣3x3y2
=﹣9x2y2﹣3x3y2.
11.(闵行区校级开学)9(xy)3•(﹣ )2+(﹣x2y)2+(﹣x2y)3•xy2.
【解答】解:原式=9x3y3• x4y2+x4y2+(﹣x6y3)•xy2
=x7y5+x4y2﹣x7y5
=x4y2.
12.(阜宁县期末)计算:2m3n•(﹣3mn2)2.
【解答】解:原式=2m3n•9m2n4
=18m5n5.
13.(涪陵区校级期中)(1)x•x2•x3+(x2)3﹣2(x3)2;
(2)(﹣4am+1)3+[2(2am)2•a].
【解答】解:(1)原式=x6+x6﹣2x6
=0;
(2)原式=(﹣4)3•a3m+3+2×4a2m•a
=﹣64a3m+3+8a2m+1.
14.(雁塔区校级三模)化简:m3n•(﹣2n)3﹣(﹣mn)2•mn2.
【解答】解:原式=﹣m3n•(8n3)﹣m2n2•mn2=﹣8m3n4﹣m3n4=﹣9m3n4.
15.(龙游县月考)计算:
(1)32×(﹣3)2;
(2)2a2•a4﹣(﹣a2)3.
【解答】解:(1)原式=9×9=81;
(2)原式=2a6﹣(﹣a6)=2a6+a6=3a6.
16.(徐汇区校级月考)计算:(﹣ ab)•(﹣4a2b)+6a•(﹣2ab)2.
【解答】解:(﹣ ab)•(﹣4a2b)+6a•(﹣2ab)2
= a3b2+6a•4a2b2
= a3b2+24a3b2.
= a3b2.
17.(青浦区月考)计算:﹣2x2yz•(﹣ xy2z)•(9xyz2).
【解答】解:原式=2× ×9x2+1+1y1+2+1z1+1+2
=3x4y4z4.
18.(贺兰县期中)若xn=3,yn=4,求(2xn)2•2yn的值.
【解答】解:∵xn=3,yn=4,
∴(2xn)2•2yn
=(2×3)2×2×4
=36×2×4
=288.
19.(儋州校级月考)(1)(﹣3ab)•(﹣2a)•(﹣a2b3);
(2) .
【解答】解:(1)(﹣3ab)•(﹣2a)•(﹣a2b3)=6a2b•(﹣a2b3)=﹣6a4b4.
(2)
=2a2b4× a2b4
= a4b8