| a+2b |
| 7 |
| 3b−2c |
| 5 |
| c−2a |
| 3 |
可设
| a+2b |
| 7 |
| 3b−2c |
| 5 |
| c−2a |
| 3 |
∴a+2b=7k,3b-2c=5k c-2a=3k,
∴a=-
| k |
| 11 |
| 39 |
| 11 |
| 31 |
| 11 |
∴
| 3a+b−2c |
| 2a−5b+6c |
3×(−
| ||||||
2×(−
|
| 26 |
| 11 |
故答案为:
| 26 |
| 11 |
| a+2b |
| 7 |
| 3b−2c |
| 5 |
| c−2a |
| 3 |
| 3a+b−2c |
| 2a−5b+6c |
| a+2b |
| 7 |
| 3b−2c |
| 5 |
| c−2a |
| 3 |
| a+2b |
| 7 |
| 3b−2c |
| 5 |
| c−2a |
| 3 |
| k |
| 11 |
| 39 |
| 11 |
| 31 |
| 11 |
| 3a+b−2c |
| 2a−5b+6c |
3×(−
| ||||||
2×(−
|
| 26 |
| 11 |
| 26 |
| 11 |