> 数学 >
求下列式子的值.
(1)
4
5
×[(−
1
2
)÷(0.75−1)+(−2)5]

(2)(−3)2÷2
1
4
×(−
2
3
)2+4−22×(−
1
3
)
人气:131 ℃ 时间:2020-05-02 08:40:54
解答
(1)
4
5
×[(−
1
2
)÷(0.75−1)+(−2)5]

=-
4
5
×[(-
1
2
)÷(
3
4
-1)-32]
=-
4
5
×[(-
1
2
)÷(-
1
4
)-32]
=-
4
5
×[(-
1
2
)×(-4)-32]
=-
4
5
×[2+(-32)]
=-
4
5
×(-30)
=24;
(2)(−3)2÷2
1
4
×(−
2
3
)
2
+4−22×(−
1
3
)

=9÷
9
4
×
4
9
+4-4×(-
1
3

=9×
4
9
×
4
9
+4+
4
3

=
16
9
+
36
9
+
12
9

=
64
9
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