若a=0.5x+20,b=0.5x+19,c=0.5x+21,求a^2+b^2+c^2-ab-bc-ac的值
a^2+b^2+c^2-ab-bc-ac
=(0.5x+20)²+(0.5x+19)²+(0.5+21)²-(0.5x+20)(0.5x+19)-(0.5x+20)(0.5x+21)-(0.5x+19)(0.5x+21)
=(0.5x)²+20x+20²+(0.5x)²+19x+19²+(0.5x)²+21x+21²-(0.5x)²-10x-9.5x-380-(0.5x)²-10x-10.5x-420-(0.5x)²-9.5x-10.5x-399
=20²+19²+21²-380-420-399
=400+361+441-380-420-399
=3