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.
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