题目内容
用简便方法计算
(1)(
+
)×5×7
(2)999
+99
+9
+
(3))299÷(299+
)
(4)[
-(
+
)]×
.
(1)(
| 1 |
| 5 |
| 2 |
| 7 |
(2)999
| 8 |
| 9 |
| 8 |
| 9 |
| 8 |
| 9 |
| 1 |
| 3 |
(3))299÷(299+
| 299 |
| 300 |
(4)[
| 13 |
| 8 |
| 5 |
| 8 |
| 5 |
| 7 |
| 3 |
| 4 |
分析:(1)运用乘法分配律解答,
(2)化
=
,然后把带分数化为整数加分数的形式,最后运用加法交换律和结合律即可解答,
(3)把带分数化为整数加分数的形式,再运用加法结合律以及乘法分配律即可解答.
(4)先运用减法性质化简算式,再依据四则运算顺序解答.
(2)化
| 1 |
| 3 |
| 3 |
| 9 |
(3)把带分数化为整数加分数的形式,再运用加法结合律以及乘法分配律即可解答.
(4)先运用减法性质化简算式,再依据四则运算顺序解答.
解答:解:(1)(
+
)×5×7,
=
×5×7+
×5×7,
=7+10,
=17;
(2)999
+99
+9
+
,
=999+99+9+(
×3+
),
=999+99+9+
×(8+1),
=999+99+9+3
=1110;
(3))299÷(299+
),
=299÷[299×(1+
)],
=299÷299÷
,
=
;
(4)[
-(
+
)]×
,
=[
-
-
)]×
,
=[1-
]×
,
=
×
,
=
.
| 1 |
| 5 |
| 2 |
| 7 |
=
| 1 |
| 5 |
| 2 |
| 7 |
=7+10,
=17;
(2)999
| 8 |
| 9 |
| 8 |
| 9 |
| 8 |
| 9 |
| 1 |
| 3 |
=999+99+9+(
| 8 |
| 9 |
| 1 |
| 3 |
=999+99+9+
| 1 |
| 3 |
=999+99+9+3
=1110;
(3))299÷(299+
| 299 |
| 300 |
=299÷[299×(1+
| 1 |
| 300 |
=299÷299÷
| 301 |
| 300 |
=
| 300 |
| 301 |
(4)[
| 13 |
| 8 |
| 5 |
| 8 |
| 5 |
| 7 |
| 3 |
| 4 |
=[
| 13 |
| 8 |
| 5 |
| 8 |
| 5 |
| 7 |
| 3 |
| 4 |
=[1-
| 5 |
| 7 |
| 3 |
| 4 |
=
| 2 |
| 7 |
| 3 |
| 4 |
=
| 3 |
| 14 |
点评:正确选择并运用简便算法是解答本题的关键,注意计算结果的准确性.
练习册系列答案
相关题目