题目内容
(1)(
+
+
)÷
+(
+
+
)÷
;
(2)2014×(1+0.5)×(1+
)×(1+0.25)×…×(1+
);
(3)
=
.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 30 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 105 |
(2)2014×(1+0.5)×(1+
| 1 |
| 3 |
| 1 |
| 99 |
(3)
| 3(y+1) |
| 5 |
| 4(y+3) |
| 7 |
考点:分数的巧算,解比例
专题:计算问题(巧算速算)
分析:(1)用过观察,此题如果按部就班地进行会很麻烦,因此可以运用设数法解决.
(2)把小数化为分数,然后计算出各个括号内的结果,约分计算.
(3)根据比例的基本性质进行解答.
(2)把小数化为分数,然后计算出各个括号内的结果,约分计算.
(3)根据比例的基本性质进行解答.
解答:
解:(1)设
+
=a,得
(
+
+
)÷
+(
+
+
)÷
=(
+a)×30+(a+
)×105
=15+30a+105a+15
=30+135a
=30+135×(
+
)
=30+135×
+135×
=30+45+27
=102
(2)2014×(1+0.5)×(1+
)×(1+0.25)×…×(1+
)
=2014×(1+
)×(1+
)×(1+
)×…×(1+
)
=2014×
×
×
×…×
=2014×
×100
=100700
(3)
=
3(y+1)×7=5×4(y+3)
21y+21=20y+60
y=39
| 1 |
| 3 |
| 1 |
| 5 |
(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 30 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 105 |
=(
| 1 |
| 2 |
| 1 |
| 7 |
=15+30a+105a+15
=30+135a
=30+135×(
| 1 |
| 3 |
| 1 |
| 5 |
=30+135×
| 1 |
| 3 |
| 1 |
| 5 |
=30+45+27
=102
(2)2014×(1+0.5)×(1+
| 1 |
| 3 |
| 1 |
| 99 |
=2014×(1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 99 |
=2014×
| 3 |
| 2 |
| 4 |
| 3 |
| 5 |
| 4 |
| 100 |
| 99 |
=2014×
| 1 |
| 2 |
=100700
(3)
| 3(y+1) |
| 5 |
| 4(y+3) |
| 7 |
3(y+1)×7=5×4(y+3)
21y+21=20y+60
y=39
点评:此题主要考查学生能否根据数字特点,通过转化的数学思想,巧妙灵活地运用运算定律,使复杂的问题简单化.
练习册系列答案
相关题目