题目内容
计算
(1)(
ab2-2ab)•
ab;
(2)-2x•(
x2y+3y-1).
(1)(
| 2 |
| 3 |
| 1 |
| 2 |
(2)-2x•(
| 1 |
| 2 |
分析:(1)先去括号,然后计算乘法,再合并同类项;
(2)先去括号,然后计算乘法,再合并同类项.
(2)先去括号,然后计算乘法,再合并同类项.
解答:解:(1)(
ab2-2ab)•
ab
=
ab2•
ab-2ab•
ab,
=
a2b3-a2b2;
(2)-2x•(
x2y+3y-1)
=-2x•
x2y+(-2x)•3y-(-2x)•1,
=-x3y+(-6xy)-(-2x)
=-x3y-6xy+2x.
| 2 |
| 3 |
| 1 |
| 2 |
=
| 2 |
| 3 |
| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 3 |
(2)-2x•(
| 1 |
| 2 |
=-2x•
| 1 |
| 2 |
=-x3y+(-6xy)-(-2x)
=-x3y-6xy+2x.
点评:本题考查了单项式乘多项式.单项式与多项式相乘的运算法则:单项式与多项式相乘,就是用单项式去乘多项式的每一项,再把所得的积相加.
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