第1个回答 2011-06-19
解:a²+b²+c²-ab-bc-ac
=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a-b)²+(b-c)²+(c-a)²]/2
=[(2006-2007)²+(2007-2008)²+(2008-2006)²]/2
=3本回答被提问者采纳
第2个回答 2011-06-19
a²+b²+c²-ab-bc-ac
=[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)]*1/2
=[(a-b)²+(b-c)²+(a-c)²]*1/2
=[(-1)²+(-1)²+(-2)²]*1/2
=3
第3个回答 2011-06-19
2006x2006+2007x2007+2008x2008-2006x2007-2007x2008-2006x2008
=2007x(2007-2006)+2008x(2008-2007)-2006x(2008-2006)
=2007+2008-4012
=3