题目内容
脱式计算
| 2-
| (15-14×
| ||||||||||||||||||
102×
|
|
|
考点:分数的四则混合运算
专题:运算顺序及法则
分析:(1)(4)运用乘法分配律解答,
(2)运用减法性质解答,
(3)按照先算括号里面的乘法,再算括号里面的减法,最后算括号外面的乘法顺序计算解答,
(5)(6)按照先算小括号里面的,再算中括号里面的,最后算括号外面的顺序计算解答.
(2)运用减法性质解答,
(3)按照先算括号里面的乘法,再算括号里面的减法,最后算括号外面的乘法顺序计算解答,
(5)(6)按照先算小括号里面的,再算中括号里面的,最后算括号外面的顺序计算解答.
解答:
解:(1)
÷9+
×
=
×(
+
)
=
×1
=
;
(2)2-
÷
-
=2-
-
=2-(
+
)
=2-2
=0;
(3)(15-14×
)×
=(15-8)×
=7×
=2
;
(4)102×
=101×
+1×
=3+
=3
;
(5)
÷[(
+
)×
]
=
÷[
×
]
=
÷
=3;
(6)
÷[
×(
+
)]
=
÷[
×
]
=
÷
=
.
| 5 |
| 17 |
| 1 |
| 9 |
| 12 |
| 17 |
=
| 1 |
| 9 |
| 5 |
| 17 |
| 12 |
| 17 |
=
| 1 |
| 9 |
=
| 1 |
| 9 |
(2)2-
| 6 |
| 13 |
| 9 |
| 26 |
| 2 |
| 3 |
=2-
| 4 |
| 3 |
| 2 |
| 3 |
=2-(
| 4 |
| 3 |
| 2 |
| 3 |
=2-2
=0;
(3)(15-14×
| 4 |
| 7 |
| 8 |
| 21 |
=(15-8)×
| 8 |
| 21 |
=7×
| 8 |
| 21 |
=2
| 2 |
| 3 |
(4)102×
| 3 |
| 101 |
=101×
| 3 |
| 101 |
| 3 |
| 101 |
=3+
| 3 |
| 101 |
=3
| 3 |
| 101 |
(5)
| 4 |
| 5 |
| 1 |
| 3 |
| 2 |
| 5 |
| 4 |
| 11 |
=
| 4 |
| 5 |
| 11 |
| 15 |
| 4 |
| 11 |
=
| 4 |
| 5 |
| 4 |
| 15 |
=3;
(6)
| 9 |
| 20 |
| 1 |
| 2 |
| 2 |
| 3 |
| 4 |
| 5 |
=
| 9 |
| 20 |
| 1 |
| 2 |
| 22 |
| 15 |
=
| 9 |
| 20 |
| 11 |
| 15 |
=
| 27 |
| 44 |
点评:本题主要考查学生依据四则运算计算方法正确进行计算,以及正确运用简便方法解决问题的能力.
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