问题 解答题
(1)计算:-
3
2
×[-32×(-
2
3
)2+(-2)3]

(2)23×(-5)-(-3)÷
3
128

(3)化简求值:5(3a2b-ab2)-(ab2+3a2b),其中a=
1
2
,b=
1
3
答案

(1)-

3
2
×[-32×(-
2
3
)2+(-2)3],

=-

3
2
×[-9×
4
9
+(-8)],

=-

3
2
×(-12),

=18;

(2)23×(-5)-(-3)÷

3
128

=-115+3×

128
3

=115+128,

=243;

(3)5(3a2b-ab2)-(ab2+3a2b),

=15a2b-5ab2-ab2-3a2b,

=12a2b-6ab2

=6ab(2a-b),

a=

1
2
,b=
1
3
代入计算得:

6ab(2a-b),

=6×

1
2
×
1
3
(2×
1
2
-
1
3
),

=

2
3

故5(3a2b-ab2)-(ab2+3a2b)=

2
3

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