已知a=-2012,b=2011,c=-2010,求a²+b²+c²+ab+bc-ac的值

a²+b²+c²+ab+bc-ac
=1/2(2a²+2b²+2c²+2ab+2bc-2ac)
=1/2(a²+2ab+b²+b²+2bc+c²+2bc+a²-2ac+c²)
=1/2[(a+b)²+(b+c)²+(a-c)²]
=1/2[(-2012+2011)²+(2011-2010)²+(-2012+2010)²]
=1/2[(-1)²+1²+(-2)²]
=1/2[1+1+4]
=1/2×6
=3