题目内容
(2013?东城区模拟)
| 计算,能简便计算的要简便计算. (1)(9+92+93)×0.01 |
(2)13.5×[1.5×(1.07+1.93)] | (3)
| ||||||||||||||||
(4)(333+667)÷[63×(
|
(5)
|
(5)1998÷1998
|
分析:算式(1)括号中的可根据乘法分配律计算;
算式(2)(3)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;
算式(4)中括号内的可根据乘法分配律计算;
算式(5)可根据
=
-
进行计算;
算式(6)可将带分数化为假分数后通过约分计算.
算式(2)(3)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;
算式(4)中括号内的可根据乘法分配律计算;
算式(5)可根据
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
算式(6)可将带分数化为假分数后通过约分计算.
解答:解:(1)(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;
(2)13.5×[1.5×(1.07+1.93)]
=13.5×[1.5×3],
=13.5×4.5,
=60.75;
(3)
+
×(
-
)
=
+
×(
-
),
=
+
×
,
=
+
,
=
+
,
=
;
(4)(333+667)÷[63×(
-
)]
=1000÷[63×
-63×
],
=1000÷[36-28],
=1000÷8,
=125;
(5)
+
+
+
+
=
+
+
+
+
,
=
-
+
-
+
-
+
-
+
-
,
=
-
,
=
;
(5)1998÷1998
,
=1998÷
,
=1998×
,
=
.
=[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;
(2)13.5×[1.5×(1.07+1.93)]
=13.5×[1.5×3],
=13.5×4.5,
=60.75;
(3)
| 1 |
| 12 |
| 11 |
| 12 |
| 5 |
| 2 |
| 1 |
| 3 |
=
| 1 |
| 12 |
| 11 |
| 12 |
| 15 |
| 6 |
| 2 |
| 6 |
=
| 1 |
| 12 |
| 11 |
| 12 |
| 13 |
| 6 |
=
| 1 |
| 12 |
| 143 |
| 72 |
=
| 6 |
| 72 |
| 143 |
| 72 |
=
| 149 |
| 72 |
(4)(333+667)÷[63×(
| 4 |
| 7 |
| 4 |
| 9 |
=1000÷[63×
| 4 |
| 7 |
| 4 |
| 9 |
=1000÷[36-28],
=1000÷8,
=125;
(5)
| 1 |
| 30 |
| 1 |
| 42 |
| 1 |
| 56 |
| 1 |
| 72 |
| 1 |
| 90 |
=
| 1 |
| 5×6 |
| 1 |
| 6×7 |
| 1 |
| 7×8 |
| 1 |
| 8×9 |
| 1 |
| 9×10 |
=
| 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 |
(5)1998÷1998
| 1998 |
| 1999 |
=1998÷
| 1998×1999+1998 |
| 1999 |
=1998×
| 1999 |
| 1998×(1999+1) |
=
| 1999 |
| 2000 |
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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