若a=1/120x+20,b=1/120+19,c=1/120+21,求代数式a^2+b^2+c^2-ab-bc-ac的

a-b=1,b-c=-2,c-a=1
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+4+1)/2
=3