题目内容
递等式计算(能简便的要简便计算)
(1)
÷(
+
)
(2)
+(
-
)÷
(3)
×[
-(
-
)].
(1)
| 13 |
| 16 |
| 7 |
| 8 |
| 13 |
| 16 |
(2)
| 7 |
| 8 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 4 |
(3)
| 4 |
| 9 |
| 3 |
| 4 |
| 7 |
| 16 |
| 1 |
| 4 |
分析:算式(1)(2)根据四则混合运算的运算顺序计算即可:选算乘除,再算加减,有括号的要先算括号里面的.
算式(3)中括号里面的减法算式可根据一个数减去两个数的差等于用这个数减去两个数中的被减数加上减数的减法性质及交换律进行计算.
算式(3)中括号里面的减法算式可根据一个数减去两个数的差等于用这个数减去两个数中的被减数加上减数的减法性质及交换律进行计算.
解答:解:(1)
÷(
+
)
=
÷(
+
),
=
÷
,
=
×
,
=
;
(2)
+(
-
)÷
=
+(
-
)×4,
=
+
×4,
=
+
,
=
+
,
=
;
(3)
×[
-(
-
)]
=
×[
+
-
],
=
×
,
=
.
| 13 |
| 16 |
| 7 |
| 8 |
| 13 |
| 16 |
=
| 13 |
| 16 |
| 14 |
| 16 |
| 13 |
| 16 |
=
| 13 |
| 16 |
| 27 |
| 16 |
=
| 13 |
| 16 |
| 16 |
| 27 |
=
| 13 |
| 27 |
(2)
| 7 |
| 8 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 4 |
=
| 7 |
| 8 |
| 5 |
| 20 |
| 4 |
| 20 |
=
| 7 |
| 8 |
| 1 |
| 20 |
=
| 7 |
| 8 |
| 1 |
| 5 |
=
| 35 |
| 40 |
| 8 |
| 40 |
=
| 43 |
| 40 |
(3)
| 4 |
| 9 |
| 3 |
| 4 |
| 7 |
| 16 |
| 1 |
| 4 |
=
| 4 |
| 9 |
| 3 |
| 4 |
| 1 |
| 4 |
| 7 |
| 16 |
=
| 4 |
| 9 |
| 9 |
| 16 |
=
| 1 |
| 4 |
点评:完成有关于分数的四则混合运算关键要细心,注意通分约分.
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