题目内容
计算
(1)8.01+99×8.01
(2)125×(8+0.8+0.08+0.008)
(3)
+
+
+
+
(4)(0.4+
)÷(
-0.75)×0.5
(5)989898×999999÷10101÷111111.
(1)8.01+99×8.01
(2)125×(8+0.8+0.08+0.008)
(3)
1 |
2 |
1 |
4 |
1 |
8 |
1 |
16 |
1 |
32 |
(4)(0.4+
2 |
3 |
5 |
6 |
(5)989898×999999÷10101÷111111.
分析:(1)(2)运用乘法分配律简算;
(3)通过观察,每个分数都可以拆成两个分数相减的形式,然后通过加减相抵消的方法,求得结果;
(4)先算括号内的,原式变为
÷
×
,把除法变为乘法,约分即可;
(5)把989898看作98×10101,把999999看作9×111111,然后调整一下运算顺序,使运算简便.
(3)通过观察,每个分数都可以拆成两个分数相减的形式,然后通过加减相抵消的方法,求得结果;
(4)先算括号内的,原式变为
16 |
15 |
1 |
12 |
1 |
2 |
(5)把989898看作98×10101,把999999看作9×111111,然后调整一下运算顺序,使运算简便.
解答:解:(1)8.01+99×8.01,
=(1+99)×8.01,
=100×8.01,
=801;
(2)125×(8+0.8+0.08+0.008),
=125×8+125×0.8+125×0.08+125×0.008,
=1000+100+10+1,
=1111;
(3)
+
+
+
+
,
=1-
+
-
+
-
+
-
+
-
,
=1-
,
=
;
(4)(0.4+
)÷(
-0.75)×0.5,
=(
+
)÷(
-
)×0.5,
=
÷
×
,
=
×12×
,
=
;
(5)989898×999999÷10101÷111111,
=(98×10101)×(9×111111)÷10101÷111111,
=(98×10101÷10101)×(9×111111÷111111),
=98×9,
=(100-2)×9,
=900-18,
=882.
=(1+99)×8.01,
=100×8.01,
=801;
(2)125×(8+0.8+0.08+0.008),
=125×8+125×0.8+125×0.08+125×0.008,
=1000+100+10+1,
=1111;
(3)
1 |
2 |
1 |
4 |
1 |
8 |
1 |
16 |
1 |
32 |
=1-
1 |
2 |
1 |
2 |
1 |
4 |
1 |
4 |
1 |
8 |
1 |
8 |
1 |
16 |
1 |
16 |
1 |
32 |
=1-
1 |
32 |
=
31 |
32 |
(4)(0.4+
2 |
3 |
5 |
6 |
=(
2 |
5 |
2 |
3 |
5 |
6 |
3 |
4 |
=
16 |
15 |
1 |
12 |
1 |
2 |
=
16 |
15 |
1 |
2 |
=
32 |
5 |
(5)989898×999999÷10101÷111111,
=(98×10101)×(9×111111)÷10101÷111111,
=(98×10101÷10101)×(9×111111÷111111),
=98×9,
=(100-2)×9,
=900-18,
=882.
点评:完成本题要注意分析式中数据,运用合适的简便方法计算.
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