问题 解答题
计算
(1)(-2ab23
(2)(3a2b-ab2
1
2
ab

(3)
327
-(3.14-π)0
(4)|
2
-
3
|+2
2
答案

(1)(-2ab23=(-2)3a3(b23=-8a3b6

(2)原式=3a2

1
2
ab-ab2÷
1
2
ab=6a-2b;                    

(3)原式=3-1=2;

(4)原式=

3
-
2
+2
2
=
3
+
2

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