题目内容
在下面各数中间,填上适当的运算符号和括号,使等式成立.
=0;
=1.
| 1 |
| 2 |
+
+
| 1 |
| 2 |
-
-
| 1 |
| 2 |
-
-
| 1 |
| 2 |
| 1 |
| 2 |
+
+
| 1 |
| 2 |
+
+
| 1 |
| 2 |
+
+
| 1 |
| 2 |
)×
)×
| 1 |
| 2 |
分析:本题可结合式中的数据根据四则混合运算的运算顺序进行分析,添上适当的运算符号及括号使等式成立:
(1)因为
+
=1,1-
=
,
-
=0,据此即可解答;
(2)因为2×
=1,前面的4个
相加的和正好是2,据此即可解答.
(1)因为
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| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
(2)因为2×
| 1 |
| 2 |
| 1 |
| 2 |
解答:解:
+
-
-
=0
(
+
+
+
)×
=2×
=1.
故答案为:+,-,-;(,+,+,+,)×.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
(
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=2×
| 1 |
| 2 |
=1.
故答案为:+,-,-;(,+,+,+,)×.
点评:完成此类题目,可在分析数据的基础上试着添加计算,最后得出正确答案.
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