题目内容
|
| ||||||||||||
|
20.07×39+200.7×4.1+40×10.035 | ||||||||||||
| 2012-2011+2010-2009+2008-2007+…+4-3+2-1. |
分析:(1)把除法改为乘法,括号内运用乘法分配律简算;
(2)把分子中的997看作1998-1,运用乘法分配律简算,进过计算,发现分子与分母相同,因此.得数为;
(3)把每个分数的分母写成两个自然数相乘的形式,然后通过拆分,简算即可;
(4)把原式变为20.07×39+20.07×41+20.07×20,运用乘法分配律简算;
(5)通过观察,每两个数分为一组,每组的结果为1,共分成2012÷2组,计算即可.
(2)把分子中的997看作1998-1,运用乘法分配律简算,进过计算,发现分子与分母相同,因此.得数为;
(3)把每个分数的分母写成两个自然数相乘的形式,然后通过拆分,简算即可;
(4)把原式变为20.07×39+20.07×41+20.07×20,运用乘法分配律简算;
(5)通过观察,每两个数分为一组,每组的结果为1,共分成2012÷2组,计算即可.
解答:解:(1)
×(4.85÷
-3.6+6.15×3
),
=
×(4.85×
-3.6+6.15×
),
=
×[(4.85-1+6.15)×
],
=
×[10×
],
=
×36,
=28;
(2)
,
=
,
=
,
=
,
=1;
(3)
+
+
+
+
+
,
=
+
+
+
+
+
,
=
×[(1-
)+(
-
)+(
-
)+(
-
)+(
-
)+(
-
)],
=
×[1-
],
=
×
,
=
;
(4)20.07×39+200.7×4.1+40×10.035,
=20.07×39+20.07×41+2×10.035×20,
=20.07×39+20.07×41+20.07×20,
=20.07×(39+41+20),
=20.07×100,
=2007;
(5)2012-2011+2010-2009+2008-2007+…+4-3+2-1,
=(2012-2011)+(2010-2009)+…+(2-1),
=1×(2012÷2),
=1006.
| 7 |
| 9 |
| 5 |
| 18 |
| 3 |
| 5 |
=
| 7 |
| 9 |
| 18 |
| 5 |
| 18 |
| 5 |
=
| 7 |
| 9 |
| 18 |
| 5 |
=
| 7 |
| 9 |
| 18 |
| 5 |
=
| 7 |
| 9 |
=28;
(2)
| 1998+1997×1999 |
| 1998×1999-1 |
=
| 1998+(1998-1)×1999 |
| 1998×1999-1 |
=
| 1998+1998×1999-1999 |
| 1998×1999-1 |
=
| 1998×1999-1 |
| 1998×1999-1 |
=1;
(3)
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
| 1 |
| 143 |
=
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 11 |
| 1 |
| 13 |
=
| 1 |
| 2 |
| 1 |
| 13 |
=
| 1 |
| 2 |
| 12 |
| 13 |
=
| 6 |
| 13 |
(4)20.07×39+200.7×4.1+40×10.035,
=20.07×39+20.07×41+2×10.035×20,
=20.07×39+20.07×41+20.07×20,
=20.07×(39+41+20),
=20.07×100,
=2007;
(5)2012-2011+2010-2009+2008-2007+…+4-3+2-1,
=(2012-2011)+(2010-2009)+…+(2-1),
=1×(2012÷2),
=1006.
点评:完成此题,注意分析式中数据,运用运算技巧和运算定律,灵活简算.
练习册系列答案
相关题目