题目内容
用递递等式计算.
4÷
-
÷4
105×13-270÷18
÷[(
+
)×
].
4÷
| 4 |
| 5 |
| 4 |
| 5 |
105×13-270÷18
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 5 |
| 1 |
| 13 |
(1)4÷
-
÷4,
=4×
-
×
,
=5-
,
=
;
(2)105×13-270÷18
=1365-15,
=1350;
(3)
÷[(
+
)×
],
=
÷[(
+
)×
],
=
÷[
×
],
=
÷
,
=
×15,
=5.
| 4 |
| 5 |
| 4 |
| 5 |
=4×
| 5 |
| 4 |
| 4 |
| 5 |
| 1 |
| 4 |
=5-
| 1 |
| 5 |
=
| 24 |
| 5 |
(2)105×13-270÷18
=1365-15,
=1350;
(3)
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 5 |
| 1 |
| 13 |
=
| 1 |
| 3 |
| 10 |
| 15 |
| 3 |
| 15 |
| 1 |
| 13 |
=
| 1 |
| 3 |
| 13 |
| 15 |
| 1 |
| 13 |
=
| 1 |
| 3 |
| 1 |
| 15 |
=
| 1 |
| 3 |
=5.
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