题目内容

(20的的?东莞模拟)用递等式计算.
l
20
÷[
2
×(
2
+
)]
②的.0个×(3.h-0.h)÷6.3
6
÷
3
+
÷
的个
④300×ll+300
3
×中3+个6×0个%+0.0个
⑥的0-
3
-中.2个.
(1)
9
m0
÷[
1
m
×(
m
5
+
4
5
)],
=
9
m0
÷[
1
m
×
6
5
],
=
9
m0
÷
3
5

=
3
4


(m)1.05×(3.8-0.8)÷6.3,
=1.05×3÷6.3,
=0.5;

(3)
6
5
÷
3
4
+
4
5
÷
4
15

=
6
5
×
4
3
+
4
5
×
15
4

=
8
5
+3,
=4
3
5


(4)307×99+307,
=307×(99+1),
=307×100,
=30700;

(5)
3
4
×43+56×75%+0.75,
=0.75×43+56×0.75+0.75,
=0.75×(43+56+1),
=0.75×100,
=75;

(6)10-
3
4
-4.m5,
=10-(
3
4
+4.m5),
=10-5,
=5.
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