题目内容
已知
+
=
,
+
=
,
+
=
,求
+
+
的值.
| 1 |
| x |
| 1 |
| y+z |
| 1 |
| 2 |
| 1 |
| y |
| 1 |
| z+x |
| 1 |
| 3 |
| 1 |
| z |
| 1 |
| x+y |
| 1 |
| 4 |
| 2 |
| x |
| 3 |
| y |
| 4 |
| z |
考点:对称式和轮换对称式
专题:
分析:由
+
=
,
+
=
,
+
=
,易得
=
,
=
,
=
,然后代入即可求得答案.
| 1 |
| x |
| 1 |
| y+z |
| 1 |
| 2 |
| 1 |
| y |
| 1 |
| z+x |
| 1 |
| 3 |
| 1 |
| z |
| 1 |
| x+y |
| 1 |
| 4 |
| 1 |
| x |
| y+z |
| 2(x+y+z) |
| 1 |
| y |
| z+x |
| 3(x+y+z) |
| 1 |
| z |
| x+y |
| 4(x+y+z) |
解答:解:∵
+
=
,
∴
=
,
∴x(y+z)=2(x+y+z),
∴x=
,
即:
=
,
同理:
=
,
=
,
∴
+
+
=
+
+
=
+
+
=
=2.
| 1 |
| x |
| 1 |
| y+z |
| 1 |
| 2 |
∴
| x+y+z |
| x(y+z) |
| 1 |
| 2 |
∴x(y+z)=2(x+y+z),
∴x=
| 2(x+y+z) |
| y+z |
即:
| 1 |
| x |
| y+z |
| 2(x+y+z) |
同理:
| 1 |
| y |
| z+x |
| 3(x+y+z) |
| 1 |
| z |
| x+y |
| 4(x+y+z) |
∴
| 2 |
| x |
| 3 |
| y |
| 4 |
| z |
| 2(y+z) |
| 2(x+y+z) |
| 3(z+x) |
| 3(x+y+z) |
| 4(x+y) |
| 4(x+y+z) |
| y+z |
| x+y+z |
| x+z |
| x+y+z |
| x+y |
| x+y+z |
| 2(x+y+z) |
| x+y+z |
点评:此题考查了对称式与轮换对称式的知识.此题难度适中,解题的关键是得到:
=
,
=
,
=
.
| 1 |
| x |
| y+z |
| 2(x+y+z) |
| 1 |
| y |
| z+x |
| 3(x+y+z) |
| 1 |
| z |
| x+y |
| 4(x+y+z) |
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