题目内容
设a=
,b=
,c=
,且x+y+z≠0,则
+
+
= .
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
| a |
| a+1 |
| b |
| b+1 |
| c |
| c+1 |
考点:对称式和轮换对称式
专题:
分析:∵a=
,b=
,c=
分别代入
,
,
表示出
,
,
的值,然后化简就可以求出结果了.
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
| a |
| a+1 |
| b |
| b+1 |
| c |
| c+1 |
| a |
| a+1 |
| b |
| b+1 |
| c |
| c+1 |
解答:解:∵a=
,b=
,c=
,
∴
=
=
=
∴
+
+
=
+
+
=
∵x+y+z≠0
∴原式=1.
故答案为:1.
| x |
| y+z |
| y |
| z+x |
| z |
| x+y |
∴
| a |
| a+1 |
| x |
| x+y+z |
| b |
| b+1 |
| y |
| x+y+z |
| c |
| c+1 |
| z |
| x+y+z |
∴
| a |
| a+1 |
| b |
| b+1 |
| c |
| c+1 |
| x |
| x+y+z |
| y |
| x+y+z |
| z |
| x+y+z |
=
| x+y+z |
| x+y+z |
∵x+y+z≠0
∴原式=1.
故答案为:1.
点评:本题是一道代数式的化简求值的题,考查了代数式的对称式和轮换对称式在化简求值中的运用.具有一定的难度.
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