题目内容
| 7 |
| 200 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 1 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 2 |
( )
( )
内应填42
42
.分析:本题可采用换元法设[
+
+
-
]为a,则原式=
×a=(
+
+
)÷(19-
),然后求出a后,即能求出缺的数据是多少.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 1 |
| 7 |
| 200 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 2 |
解答:解:设[
+
+
-
]为a,则:
×[
+
+
-
]=(
+
+
)÷(19-
)
变为
×a=(
+
+
)÷(19-
),
×a=÷
×
,
a=
×
a=
;
因为[
+
+
-
]=a
所以(
+
+
)-
=
,即这个分数的分母为42.
故答案为42.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 1 |
| 7 |
| 200 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 1 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 2 |
变为
| 7 |
| 200 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 2 |
| 7 |
| 200 |
| 37 |
| 60 |
| 2 |
| 37 |
a=
| 1 |
| 30 |
| 200 |
| 7 |
a=
| 20 |
| 21 |
因为[
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 1 |
所以(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 7 |
| 20 |
| 21 |
| 1 |
| 42 |
故答案为42.
点评:换元法是分数巧算中常用的方法之一.
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