题目内容
计算,能简便计算的要简便计算.
(1)
|
(2)1998÷1998
| ||||||||||||
| (3)(9+92+93)×0.01 | (4)13.5×[1.5×(1.07+1.93)] | ||||||||||||
(5)
|
(6)(333+667)÷[63×(
|
考点:分数的巧算,四则混合运算中的巧算
专题:计算问题(巧算速算)
分析:(1)通过仔细观察,每个分数的分母可以写成两个连续自然数相乘的积,于是把每个分数拆成两个分数相减的形式,然后通过加减相抵消的方法,得出结果.
(2)可根据=
=
-
进行计算.
(3)(6)括号中可根据乘法分配律计算.
(4)(5)根据四则混合运算顺序计算即可:先算乘除,再算加减,有括号的要先算括号的里面的.
(2)可根据=
| 1 |
| n×(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
(3)(6)括号中可根据乘法分配律计算.
(4)(5)根据四则混合运算顺序计算即可:先算乘除,再算加减,有括号的要先算括号的里面的.
解答:
解:(1)
+
+
+
+
=(
-
)+(
-
)+(
-
)+(
-
)+(
-
)
=
-
=
(2)1998÷1998
=1998÷
=1998×
=
(3)(9+92+93)×0.01
=[9+9×(9+9×9)]×0.01,
=[9+9×(9+81)]×0.01,
=[9+9×90]×0.01,
=[9+810]×0.01,
=819×0.01,
=8.19
(4)13.5×[1.5×(1.07+1.93)]
=13.5×[1.5×3]
=13.5×4.5
=60.75
(5)
+
×(
-
)
=
+
×
=
+
=
(6))(333+667)÷[63×(
-
)]
=1000÷[63×
-63×
]
=1000÷[36-28]
=1000÷8
=125
| 1 |
| 30 |
| 1 |
| 42 |
| 1 |
| 56 |
| 1 |
| 72 |
| 1 |
| 90 |
=(
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 6 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 10 |
=
| 1 |
| 5 |
| 1 |
| 10 |
=
| 1 |
| 10 |
(2)1998÷1998
| 1998 |
| 1999 |
=1998÷
| 1998×1999+1998 |
| 1999 |
=1998×
| 1999 |
| 1998×2000 |
=
| 1999 |
| 2000 |
(3)(9+92+93)×0.01
=[9+9×(9+9×9)]×0.01,
=[9+9×(9+81)]×0.01,
=[9+9×90]×0.01,
=[9+810]×0.01,
=819×0.01,
=8.19
(4)13.5×[1.5×(1.07+1.93)]
=13.5×[1.5×3]
=13.5×4.5
=60.75
(5)
| 1 |
| 12 |
| 11 |
| 12 |
| 5 |
| 2 |
| 1 |
| 3 |
=
| 1 |
| 12 |
| 11 |
| 12 |
| 13 |
| 6 |
=
| 1 |
| 12 |
| 143 |
| 72 |
=
| 149 |
| 72 |
(6))(333+667)÷[63×(
| 4 |
| 7 |
| 4 |
| 9 |
=1000÷[63×
| 4 |
| 7 |
| 4 |
| 9 |
=1000÷[36-28]
=1000÷8
=125
点评:完成本题要认真分析式中数的居特点及内在联系,然后运用合适的方法进行计算.
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