题目内容
脱式计算
(1)(6
-3
)÷(13+11
)
(2)
+
+
+
+
(3)2.25×
+2.75÷1
+60%
(4)(1+3+5+…+1001)-(2+4+6+…+1000)
(1)(6
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 5 |
(2)
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 30 |
| 1 |
| 42 |
| 1 |
| 56 |
(3)2.25×
| 3 |
| 5 |
| 2 |
| 3 |
(4)(1+3+5+…+1001)-(2+4+6+…+1000)
分析:(1)按运算顺序计算,两个括号同时计算,最后算除法;
(2)算式中的每个分数都可以拆成两个分数相减的行式,然后通过加减相抵消的方法,得出结果;
(3)把百分数化为分数,把除法改为乘法,运用乘法分配律的逆运算简算;
(4)通过仔细观察,第一个括号内的数除1以外,剩下的数与第二个括号内的数分别组合,构成500个结果为1的算式,即原式=1+(3-2)+(5-4)+…+(1001-1000),计算即可.
(2)算式中的每个分数都可以拆成两个分数相减的行式,然后通过加减相抵消的方法,得出结果;
(3)把百分数化为分数,把除法改为乘法,运用乘法分配律的逆运算简算;
(4)通过仔细观察,第一个括号内的数除1以外,剩下的数与第二个括号内的数分别组合,构成500个结果为1的算式,即原式=1+(3-2)+(5-4)+…+(1001-1000),计算即可.
解答:解:(1)(6
-3
)÷(13+11
),
=(6
-3
)÷24
,
=3
÷
,
=
×
,
=
;
(2)
+
+
+
+
,
=
-
+
-
+
-
+
-
+
-
+
-
,
=
-
,
=
;
(3)2.25×
+2.75÷1
+60%,
=2.25×
+2.75×
+1×
,
=(2.25+2.75+1)×
,
=6×
,
=3
;
(4)(1+3+5+…+1001)-(2+4+6+…+1000),
=1+(3-2)+(5-4)+…+(1001-1000),
=1+1+1+…+1,
=501.
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 5 |
=(6
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 5 |
=3
| 1 |
| 4 |
| 121 |
| 5 |
=
| 13 |
| 4 |
| 5 |
| 121 |
=
| 65 |
| 484 |
(2)
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 30 |
| 1 |
| 42 |
| 1 |
| 56 |
=
| 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 |
| 2 |
| 1 |
| 8 |
=
| 3 |
| 8 |
(3)2.25×
| 3 |
| 5 |
| 2 |
| 3 |
=2.25×
| 3 |
| 5 |
| 3 |
| 5 |
| 3 |
| 5 |
=(2.25+2.75+1)×
| 3 |
| 5 |
=6×
| 3 |
| 5 |
=3
| 3 |
| 5 |
(4)(1+3+5+…+1001)-(2+4+6+…+1000),
=1+(3-2)+(5-4)+…+(1001-1000),
=1+1+1+…+1,
=501.
点评:简便计算主要是运用所学性质与定律以及数与数之间的特殊关系灵活进行,因此应注意审题,多做几方面试探,以求得简便的算法.
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