即a-b=-1b-c=-1c-a=2所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2=[(a-b)²+(b-c)²+(c-a)²]/2=(1+1+4)/2=3
