题目内容
计算下面各题:
(1)20.02×67+3.2×200.2+400.4×0.05
(2)
+
+
+
+
.
(1)20.02×67+3.2×200.2+400.4×0.05
(2)
1 |
3 |
1 |
15 |
1 |
35 |
1 |
63 |
1 |
99 |
分析:(1)根据题意,20.02×67=200.2×6.7,400.4×0.05=200.2×2×0.05=200.2×0.1,再根据乘法分配律进行巧算即可;
(2)根据分数的拆项,把
、
、
、
、
,分别拆成
×(1-
)、
×(
-
)、
×(
-
)、
×(
-
)、
×(
-
),再进一步解答即可.
(2)根据分数的拆项,把
1 |
3 |
1 |
15 |
1 |
35 |
1 |
63 |
1 |
99 |
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
2 |
1 |
7 |
1 |
9 |
1 |
2 |
1 |
9 |
1 |
11 |
解答:解:
(1)20.02×67+3.2×200.2+400.4×0.05
=200.2×6.7+3.2×200.2+200.2×2×0.05
=200.2×6.7+3.2×200.2+200.2×0.1
=200.2×(6.7+3.2+0.1)
=200.2×10
=2002,
(2)
+
+
+
+
=
×(1-
)+
×(
-
)+
×(
-
)+
×(
-
)+
×(
-
)
=
×(1-
+
-
+
-
+
-
+
-
)
=
×(1-
)
=
×
=
.
(1)20.02×67+3.2×200.2+400.4×0.05
=200.2×6.7+3.2×200.2+200.2×2×0.05
=200.2×6.7+3.2×200.2+200.2×0.1
=200.2×(6.7+3.2+0.1)
=200.2×10
=2002,
(2)
1 |
3 |
1 |
15 |
1 |
35 |
1 |
63 |
1 |
99 |
=
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
2 |
1 |
7 |
1 |
9 |
1 |
2 |
1 |
9 |
1 |
11 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
7 |
1 |
9 |
1 |
9 |
1 |
11 |
=
1 |
2 |
1 |
11 |
=
1 |
2 |
10 |
11 |
=
5 |
11 |
点评:第一小题主要考查乘法分配律的灵活多变的一种形式,找准相同的因数,就比较容易解答此类问题;第二小题主要考查分数的拆项,由
=
×(
-
),进行解答即可.
1 |
(2n-1)(2n+1) |
1 |
2 |
1 |
2n-1 |
1 |
2n+1 |
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