题目内容
(1)200-199+198-197+196 …+2-1
(2)2012×20132013-2013×20122012
(3)1.25×98+12
×
+1
(4)
+
+
+…+
(5)0.35÷0.125÷64÷0.5÷0.25
(6)
+
-
+
-
+
-
.
(2)2012×20132013-2013×20122012
(3)1.25×98+12
| 1 |
| 2 |
| 1 |
| 10 |
| 1 |
| 4 |
(4)
| 1 |
| 2007 |
| 2 |
| 2007 |
| 3 |
| 2007 |
| 2006 |
| 2007 |
(5)0.35÷0.125÷64÷0.5÷0.25
(6)
| 1 |
| 2 |
| 5 |
| 6 |
| 7 |
| 12 |
| 9 |
| 20 |
| 11 |
| 30 |
| 13 |
| 42 |
| 15 |
| 56 |
分析:(1)两个1组,分组计算即可;
(2)将2012×20132013-2013×20122012变形为2012×2013×10001-2013×2012×10001,即可求解;
(3)根据乘法分配律简便计算;
(4)根据同分母分数的计算法则和高斯求和公式计算;
(5)先化为分数,除法变为乘法,再约分计算即可;
(6)先拆分,再抵消法计算即可求解.
(2)将2012×20132013-2013×20122012变形为2012×2013×10001-2013×2012×10001,即可求解;
(3)根据乘法分配律简便计算;
(4)根据同分母分数的计算法则和高斯求和公式计算;
(5)先化为分数,除法变为乘法,再约分计算即可;
(6)先拆分,再抵消法计算即可求解.
解答:解:(1)200-199+198-197+196 …+2-1,
=(200-199)+(198-197)+(196 …+(2-1),
=1×100,
=100;
(2)2012×20132013-2013×20122012,
=2012×2013×10001-2013×2012×10001,
=0;
(3)1.25×98+12
×
+1
,
=1.25×98+1.25+1.25,
=1.25×(98+1+1),
=1.25×100,
=125;
(4)
+
+
+…+
,
=
,
=
,
=1003;
(5)0.35÷0.125÷64÷0.5÷0.25,
=
×8×
×2×4,
=
;
(6)
+
-
+
-
+
-
,
=1-
+
+
-
-
+
+
-
-
+
+
-
-
,
=1-
,
=
.
=(200-199)+(198-197)+(196 …+(2-1),
=1×100,
=100;
(2)2012×20132013-2013×20122012,
=2012×2013×10001-2013×2012×10001,
=0;
(3)1.25×98+12
| 1 |
| 2 |
| 1 |
| 10 |
| 1 |
| 4 |
=1.25×98+1.25+1.25,
=1.25×(98+1+1),
=1.25×100,
=125;
(4)
| 1 |
| 2007 |
| 2 |
| 2007 |
| 3 |
| 2007 |
| 2006 |
| 2007 |
=
| 1+2+…+2006 |
| 2007 |
=
| (1+2006)×2006÷2 |
| 2007 |
=1003;
(5)0.35÷0.125÷64÷0.5÷0.25,
=
| 7 |
| 20 |
| 1 |
| 64 |
=
| 7 |
| 20 |
(6)
| 1 |
| 2 |
| 5 |
| 6 |
| 7 |
| 12 |
| 9 |
| 20 |
| 11 |
| 30 |
| 13 |
| 42 |
| 15 |
| 56 |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 6 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 8 |
=1-
| 1 |
| 8 |
=
| 7 |
| 8 |
点评:考查了四则混合运算中的巧算,注意灵活运用运算律简便计算.
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