∴2011-a=2012-b=2013-c,
∴b=a+1,c=a+2,又abc=24,
则
a |
bc |
b |
ac |
c |
ab |
1 |
a |
1 |
b |
1 |
c |
=
a2+b2+c2 |
abc |
bc+ac+ab |
abc |
=
a2+b2+c2-bc-ac-ab |
abc |
=
a2+(a+1)2+(a+2)2-(a+1)(a+2)-a(a+2)-a(a+1) |
24 |
=
3 |
24 |
1 |
8 |
故答案为:
1 |
8 |
a |
bc |
b |
ac |
c |
ab |
1 |
a |
1 |
b |
1 |
c |
a |
bc |
b |
ac |
c |
ab |
1 |
a |
1 |
b |
1 |
c |
a2+b2+c2 |
abc |
bc+ac+ab |
abc |
a2+b2+c2-bc-ac-ab |
abc |
a2+(a+1)2+(a+2)2-(a+1)(a+2)-a(a+2)-a(a+1) |
24 |
3 |
24 |
1 |
8 |
1 |
8 |