题目内容
已知:| x+y-xy |
| x+y+2xy |
| y+z-2yz |
| y+z+3yz |
| x+z-3zx |
| x+z+4zx |
| 2 |
| x |
| 3 |
| y |
| 1 |
| z |
分析:根据已知得出
=
=
,从而得出
=
=
,进而求出即可.
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解答:解:∵
=
=
,
∴
=
=
,
∵
=
-
,
∴
=
-
,
∴
=
=
,
设
=a,
=b,
∴
=
=
,
∴
,
整理得出:
,
解得:
,
则x=-4,y=-4,
代入
=
-
,则z=-4,
故答案为:-4,-4,-4.
| x+y-xy |
| x+y+2xy |
| y+z-2yz |
| y+z+3yz |
| x+z-3zx |
| x+z+4zx |
∴
| ||||
|
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|
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|
∵
| 2 |
| x |
| 3 |
| y |
| 1 |
| z |
∴
| 1 |
| z |
| 3 |
| y |
| 2 |
| x |
∴
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|
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|
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设
| 1 |
| x |
| 1 |
| y |
∴
| a+b-1 |
| a+b+2 |
| 3a-b-2 |
| 3a-b+3 |
| 4a-2b-3 |
| 4a-2b+4 |
∴
|
整理得出:
|
解得:
|
则x=-4,y=-4,
代入
| 2 |
| x |
| 3 |
| y |
| 1 |
| z |
故答案为:-4,-4,-4.
点评:此题主要考查了分式的混合运算,由已知得出
=
=
是解决问题的关键.
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