题目内容
若| 1 |
| x |
| 2 |
| y |
| 3 |
| z |
| 3 |
| x |
| 2 |
| y |
| 1 |
| z |
| 1 |
| x |
| 1 |
| y |
| 1 |
| z |
分析:把已知中两式相加即可轻松求解.
解答:解:两式相加得,
+
+
=12,
等式两边都除以4,得
+
+
=3.
| 4 |
| x |
| 4 |
| y |
| 4 |
| z |
等式两边都除以4,得
| 1 |
| x |
| 1 |
| y |
| 1 |
| z |
点评:根据题目特点,利用整体代入是解本题的关键,也使本题求解更加简便.
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若| 1 |
| x |
| 2 |
| y |
| 3 |
| z |
| 3 |
| x |
| 2 |
| y |
| 1 |
| z |
| 1 |
| x |
| 1 |
| y |
| 1 |
| z |
| 4 |
| x |
| 4 |
| y |
| 4 |
| z |
| 1 |
| x |
| 1 |
| y |
| 1 |
| z |