题目内容
计算
(1)(-2ab2)3
(2)(3a2b-ab2)÷
ab
(3)
-(3.14-π)0
(4)|
-
|+2
.
(1)(-2ab2)3
(2)(3a2b-ab2)÷
| 1 |
| 2 |
(3)
| 3 | 27 |
(4)|
| 2 |
| 3 |
| 2 |
(1)(-2ab2)3=(-2)3a3(b2)3=-8a3b6;
(2)原式=3a2b÷
ab-ab2÷
ab=6a-2b;
(3)原式=3-1=2;
(4)原式=
-
+2
=
+
.
(2)原式=3a2b÷
| 1 |
| 2 |
| 1 |
| 2 |
(3)原式=3-1=2;
(4)原式=
| 3 |
| 2 |
| 2 |
| 3 |
| 2 |
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