题目内容
用递等式计算.
×
÷
×
40×[93÷(
+
)].
| 3 |
| 4 |
| 4 |
| 7 |
| 6 |
| 7 |
| 7 |
| 24 |
40×[93÷(
| 2 |
| 5 |
| 3 |
| 8 |
(1)
×
÷
×
,
=
÷
×
,
=
×
,
=
;
(2)40×[93÷(
+
)],
=40×[93÷
],
=40×120,
=4800.
| 3 |
| 4 |
| 4 |
| 7 |
| 6 |
| 7 |
| 7 |
| 24 |
=
| 3 |
| 7 |
| 6 |
| 7 |
| 7 |
| 24 |
=
| 1 |
| 2 |
| 7 |
| 24 |
=
| 7 |
| 48 |
(2)40×[93÷(
| 2 |
| 5 |
| 3 |
| 8 |
=40×[93÷
| 31 |
| 40 |
=40×120,
=4800.
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