题目内容
(2008?扬州)脱式计算(能用简便算法的,要写出简便过程)
24×10.5-3549÷39
÷[(
-
)×12]
2-
÷
-
÷
-
×6.
24×10.5-3549÷39
| 7 |
| 10 |
| 3 |
| 5 |
| 1 |
| 4 |
2-
| 5 |
| 8 |
| 25 |
| 28 |
| 3 |
| 10 |
| 8 |
| 15 |
| 1 |
| 6 |
| 7 |
| 15 |
分析:算式(1)可将式中24×10.5据乘法分配律变为24×(10+0.5)进行计算,使计算更加简便.
(2)先算小括号内部的,在计算中括号内的,最后计算括号外面的.
(3)本题先计算除法然后运用“一个数连续减去两个数可以用这个数减去这两个数的和”进行计算.
(4)先计算除法在运用乘法的分配律进行计算,最终把结果化成最简分数.
(2)先算小括号内部的,在计算中括号内的,最后计算括号外面的.
(3)本题先计算除法然后运用“一个数连续减去两个数可以用这个数减去这两个数的和”进行计算.
(4)先计算除法在运用乘法的分配律进行计算,最终把结果化成最简分数.
解答:解:(1)24×10.5-3549÷39,
=24×(10+0.5)-91,
=240+12-91,,
=252-91,
=161;
(2)
÷[(
-
)×12],
=
÷[(
-
)×12],
=
÷[
×12],
=
÷
,
=
×
,
=
;
(3)2-
÷
-
,
=2-
×
-
,
=2-
-
,
=2-(
+
),
=2-1,
=1;
(4)
÷
-
×6,
=
×6-
×6,
=(
-
)×6,
=
×6,
=
.
=24×(10+0.5)-91,
=240+12-91,,
=252-91,
=161;
(2)
| 7 |
| 10 |
| 3 |
| 5 |
| 1 |
| 4 |
=
| 7 |
| 10 |
| 12 |
| 20 |
| 5 |
| 20 |
=
| 7 |
| 10 |
| 7 |
| 20 |
=
| 7 |
| 10 |
| 21 |
| 5 |
=
| 7 |
| 10 |
| 5 |
| 21 |
=
| 1 |
| 6 |
(3)2-
| 5 |
| 8 |
| 25 |
| 28 |
| 3 |
| 10 |
=2-
| 5 |
| 8 |
| 28 |
| 25 |
| 3 |
| 10 |
=2-
| 7 |
| 10 |
| 3 |
| 10 |
=2-(
| 7 |
| 10 |
| 3 |
| 10 |
=2-1,
=1;
(4)
| 8 |
| 15 |
| 1 |
| 6 |
| 7 |
| 15 |
=
| 8 |
| 15 |
| 7 |
| 15 |
=(
| 8 |
| 15 |
| 7 |
| 15 |
=
| 1 |
| 15 |
=
| 2 |
| 5 |
点评:本题主要考查了运算定律的灵活运用及分数四则混合运算的能力.
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