(1)左边=a2+b2+c2﹣ab﹣bc﹣ac, =(2a2+2b2+2c2﹣2ab﹣2bc﹣2ac), =(a2﹣2ab+b2+b2﹣2bc+c2+a2﹣2ac+c2), =[(a﹣b)2+(b﹣c)2+(c﹣a)2], ∴左边=右边, 即这个等式是正确的; (2)当a=2009,b=2010,c=2011时, a2+b2+c2﹣ab﹣bc﹣ac=[(a﹣b)2+(b﹣c)2+(c﹣a)2], =[(2009﹣2010)2+(2010﹣2011)2+(2011﹣2009)2], =×(1+1+4), =3. |