题目内容

计算或化简:
(1)(2
3
+
20
)(
12
-2
5
)

(2)
9m
-
m
3
1
m
-
1
2m
m3

(3)(3
12
-2
1
3
+
48
)÷2
3

(4)sin230°+cos245°+sin60°•tan45°.

(1)(2
3
+
20
)(
12
-2
5
)

=(2
3
+
4×5
)(
4×3
-2
5
),
=(2
3
+2
5
)(2
3
-2
5
),
=(2
3
2-(2
5
2
=12-20,
=-8;

(2)
9m
-
m
3
1
m
-
1
2m
m3

=3
m
-
m
3
×
m
m
-
1
2m
×m
m

=3
m
-
m
3
-
m
2

=(3-
1
3
-
1
2
m

=
13
6
m


(3)(3
12
-2
1
3
+
48
)÷2
3

=(3×2
3
-2×
3
3
+4
3
)÷2
3

=
28
3
3
÷2
3

=
14
3


(4)sin230°+cos245°+sin60°•tan45°,
=(
1
2
2+(
2
2
2+
3
2
×1,
=
1
4
+
1
2
+
3
2

=
3
4
+
3
2

=
3+2
3
4
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