题目内容
简便运算
| 139×
| |||||||||||
2012
| 25×(
| |||||||||||
| 11×(
| |||||||||||
|
考点:分数的简便计算
专题:运算定律及简算
分析:(1)把整数部分的86看作87-1,运用乘法分配律简算.
(2)(4)(6)运用乘法分配律简算.
(3)把带分数化成假分数,分子部分不必算出来,约分计算.
(5)通过数字转化,运用乘法分配律简算.
(7)通过分数拆分,加减相互抵消,求得结果.
(2)(4)(6)运用乘法分配律简算.
(3)把带分数化成假分数,分子部分不必算出来,约分计算.
(5)通过数字转化,运用乘法分配律简算.
(7)通过分数拆分,加减相互抵消,求得结果.
解答:
解:(1)
×86
=
×(87-1)
=
×87-
=86-
=85
(2)139×
-137×
=(138+1)×
-(138-1)×
=138×
+
-138×
+
=137-1+(
+
)
=136+1
=137
(3)2012
×
=
×
=
×
=
×
=
×
=2011
(4)25×(
+
)×17
=25×17×
+
×25×17
=119+125
=244
(5)
×18+2
×17
=
×18+6×
×17
=
×(18+6×17)
=
×220
=88
(6)11×(
-
)×17
=11×17×
-
×17×11
=17-11
=6
(7)
+
+
+
+
+…+
=1-
+
-
+
-
+
-
+
-
+…+
-
=1-
=
| 86 |
| 87 |
=
| 86 |
| 87 |
=
| 86 |
| 87 |
| 86 |
| 87 |
=86-
| 86 |
| 87 |
=85
| 1 |
| 87 |
(2)139×
| 137 |
| 138 |
| 1 |
| 138 |
=(138+1)×
| 137 |
| 138 |
| 1 |
| 138 |
=138×
| 137 |
| 138 |
| 137 |
| 138 |
| 1 |
| 138 |
| 1 |
| 138 |
=137-1+(
| 137 |
| 138 |
| 1 |
| 138 |
=136+1
=137
(3)2012
| 1 |
| 2010 |
| 2010 |
| 2011 |
=
| 2012×2010+1 |
| 2010 |
| 2010 |
| 2011 |
=
| (2011+1)×2010+1 |
| 2010 |
| 2010 |
| 2011 |
=
| 2011×2010+2011 |
| 2010 |
| 2010 |
| 2011 |
=
| 2011×(2010+1) |
| 2010 |
| 2010 |
| 2011 |
=2011
(4)25×(
| 7 |
| 25 |
| 5 |
| 17 |
=25×17×
| 7 |
| 25 |
| 5 |
| 17 |
=119+125
=244
(5)
| 2 |
| 5 |
| 2 |
| 5 |
=
| 2 |
| 5 |
| 2 |
| 5 |
=
| 2 |
| 5 |
=
| 2 |
| 5 |
=88
(6)11×(
| 1 |
| 11 |
| 1 |
| 17 |
=11×17×
| 1 |
| 11 |
| 1 |
| 17 |
=17-11
=6
(7)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 5×6 |
| 1 |
| 99×100 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 99 |
| 1 |
| 100 |
=1-
| 1 |
| 100 |
=
| 99 |
| 100 |
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或运算技巧,使复杂的问题简单化.
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