题目内容
在下列括号里填上不同的整数.
(1)
=
+
=
+
(2)
=
-
=
-
(3)
=
+
+
.
(1)
| 1 |
| 24 |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
(2)
| 1 |
| 18 |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
(3)
| 7 |
| 10 |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
| 1 |
| ( ) |
分析:(1)24=3×8=4×6,所以把
分子分母同时扩大4倍或5倍,把分数拆项为两个分数的和,然后约分,即可得解;
(2)18=2×9=3×6,所以把
分子分母同时扩大2倍或5倍,把分数拆项为两个分数的差,然后约分,即可得解;
(3)7=2+5,10=2×5,所以
=
+
=
+
,然后利用
=
+
或
=
+
,即可得解.
| 1 |
| 24 |
(2)18=2×9=3×6,所以把
| 1 |
| 18 |
(3)7=2+5,10=2×5,所以
| 7 |
| 10 |
| 2 |
| 10 |
| 5 |
| 10 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 6 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| 30 |
解答:解:(1)
=
=
+
=
+
,
=
=
+
=
+
;
(2)
=
=
-
=
-
,
=
=
-
=
-
;
(3)
=
+
=
+
=
+
+
=
+
+
.
故答案为:32,96,30,120;12,36,15,90;3,5,6.
| 1 |
| 24 |
| 4 |
| 24×4 |
| 3 |
| 24×4 |
| 1 |
| 24×4 |
| 1 |
| 32 |
| 1 |
| 96 |
| 1 |
| 24 |
| 5 |
| 24×5 |
| 4 |
| 24×5 |
| 1 |
| 24×5 |
| 1 |
| 30 |
| 1 |
| 120 |
(2)
| 1 |
| 18 |
| 2 |
| 18×2 |
| 3 |
| 18×2 |
| 1 |
| 18×2 |
| 1 |
| 12 |
| 1 |
| 36 |
| 1 |
| 18 |
| 5 |
| 18×5 |
| 6 |
| 18×5 |
| 1 |
| 18×5 |
| 1 |
| 15 |
| 1 |
| 90 |
(3)
| 7 |
| 10 |
| 2 |
| 10 |
| 5 |
| 10 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 3 |
| 1 |
| 2×3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 6 |
故答案为:32,96,30,120;12,36,15,90;3,5,6.
点评:此题考查了分数的拆项,灵活应用分数分母的因数来解决此类问题,如果分母是质数,则利用
=
+
来解决问题.
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n(n+1) |
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