题目内容
(2012?广州模拟)脱式计算(能简算的要简算)
(1)(
+
-
)×24
(2)(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)
(3)2007×
(4)2.5×0.875+0.25×1.25.
(1)(
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 24 |
(2)(1+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 9 |
(3)2007×
| 2005 |
| 2006 |
(4)2.5×0.875+0.25×1.25.
分析:(1)根据数字特点,运用乘法分配律简算;
(2)通过观察发现规律,每一个假分数(除了倒数第二项,因为它后面不再有对应的了)出现以后,在后面都会出现它的倒数,(除了倒数第二项,因为它后面不再有对应的了),最后只剩下第二项
和倒数第二项
,所以原式=
×
;
(3)把2007看作(2006+1),然后运用乘法分配律简算;
(4)根据数字特点,把0.25×1.25改写成2.5×0.125,运用乘法分配律的逆运算简算.
(2)通过观察发现规律,每一个假分数(除了倒数第二项,因为它后面不再有对应的了)出现以后,在后面都会出现它的倒数,(除了倒数第二项,因为它后面不再有对应的了),最后只剩下第二项
| 1 |
| 2 |
| 10 |
| 9 |
| 1 |
| 2 |
| 10 |
| 9 |
(3)把2007看作(2006+1),然后运用乘法分配律简算;
(4)根据数字特点,把0.25×1.25改写成2.5×0.125,运用乘法分配律的逆运算简算.
解答:解:(1)(
+
-
)×24,
=
×24+
×24-
×24,
=8+6-1,
=13;
(2)(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
),
=
×
×
×
×…×
×
,
=
×
,
=
;
(3)2007×
,
=(2006+1)×
,
=2006×
+
,
=2005+
,
=2005
;
(4)2.5×0.875+0.25×1.25,
=2.5×0.875+2.5×0.125,
=(0.875+0.125)×2.5,
=1×2.5,
=2.5.
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 24 |
=
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 24 |
=8+6-1,
=13;
(2)(1+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 9 |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 10 |
| 9 |
| 8 |
| 9 |
=
| 1 |
| 2 |
| 10 |
| 9 |
=
| 5 |
| 9 |
(3)2007×
| 2005 |
| 2006 |
=(2006+1)×
| 2005 |
| 2006 |
=2006×
| 2005 |
| 2006 |
| 2005 |
| 2006 |
=2005+
| 2005 |
| 2006 |
=2005
| 2005 |
| 2006 |
(4)2.5×0.875+0.25×1.25,
=2.5×0.875+2.5×0.125,
=(0.875+0.125)×2.5,
=1×2.5,
=2.5.
点评:此题考查学生从数字特点出发,巧妙灵活地运用所学定律或性质、以及运算技巧,得以简算的能力.
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