题目内容
已知：
=
=
≠0，则
=
．
| x |
| 2 |
| y |
| 4 |
| z |
| 5 |
| x+2y-z |
| 2x-y+3z |
| 1 |
| 3 |
| 1 |
| 3 |
分析：首先设
=
=
=k，可得x=2k，y=4k，z=5k，然后代入
，即可求得答案．
| x |
| 2 |
| y |
| 4 |
| z |
| 5 |
| x+2y-z |
| 2x-y+3z |
解答：解：设
=
=
=k，
∴x=2k，y=4k，z=5k，
∴
=
=
．
故答案为：
．
| x |
| 2 |
| y |
| 4 |
| z |
| 5 |
∴x=2k，y=4k，z=5k，
∴
| x+2y-z |
| 2x-y+3z |
| 2k+8k-5k |
| 4k-4k+15k |
| 1 |
| 3 |
故答案为：
| 1 |
| 3 |
点评：此题考查了比例的性质．此题比较简单，注意设
=
=
=k，可得x=2k，y=4k，z=5k是解此题的关键．
| x |
| 2 |
| y |
| 4 |
| z |
| 5 |
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