已知a=1/20x+20,b=1/20x+19,c=1/20x+21,求代数式a?+b?+c?-ab-bc-ac的值?

Question

已知a=1/20x+20,b=1/20x+19,c=1/20x+21,求代数式a?+b?+c?-ab-bc-ac的值?

Answer 1

a-1=b→a-b=1
a+1=c→a-c=-1→c-a=1
b+2=c→b-c=-2
a^2+b^2+c^2-ab-bc-ac
=(a^2-ab)+(b^2-bc)+(c^2-ca)
=a(a-b)+b(b-c)+c(c-a)
=a-2b+c
=a-2(a-1)+(a+1)
=a-2a+2+a+1
=3