题目内容
计算:3(a2b-ab2)+[
a2b-
(a2b-ab2)]-[
ab2+
(a2b-ab2)].
| 3 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 8 |
| 3 |
原式=3a2b-3ab2+[
a2b-
a2b+
ab2]-[
ab2+
a2b-
ab2]
=3a2b-3ab2+
a2b-
a2b+
ab2-
ab2-
a2b+
ab2
=(3+
-
-
)a2b-(3-
+
-
)ab2
=
a2b-
ab2.
| 3 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 2 |
| 8 |
| 3 |
| 8 |
| 3 |
=3a2b-3ab2+
| 3 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 2 |
| 8 |
| 3 |
| 8 |
| 3 |
=(3+
| 3 |
| 2 |
| 1 |
| 3 |
| 8 |
| 3 |
| 1 |
| 3 |
| 1 |
| 2 |
| 8 |
| 3 |
=
| 3 |
| 2 |
| 1 |
| 2 |
练习册系列答案
相关题目