题目内容
1998×(
-
)+11×(
-
)-2009×(
+
)+3=
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
| 1 |
| 2009 |
| 1 |
| 11 |
| 1 |
| 1998 |
0
0
.分析:先根据加法交换律、乘法分配律将式子变形为3-(2009-1998)×
-(1998+11)×
-(2009-11)×
,再计算即可求解.
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
解答:解:1998×(
-
)+11×(
-
)-2009×(
+
)+3,
=3-(2009-1998)×
-(1998+11)×
-(2009-11)×
,
=3-11×
-2009×
-1998×
,
=3-1-1-1,
=0.
故答案为:0.
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
| 1 |
| 2009 |
| 1 |
| 11 |
| 1 |
| 1998 |
=3-(2009-1998)×
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
=3-11×
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
=3-1-1-1,
=0.
故答案为:0.
点评:考查了分数的巧算,灵活运用运算定律将式子变形为3-(2009-1998)×
-(1998+11)×
-(2009-11)×
是解题的关键.
| 1 |
| 11 |
| 1 |
| 2009 |
| 1 |
| 1998 |
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