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 |