Question

初二代数 help!已知a-b=3 b-c=2 求a^2+b^2+c^2-ab-bc-ac

Answer 1

a-b=3

b-c=2

=>a-b+b-c=a-c=5

a^2+b^2+c^2-ab-bc-ac

=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/2

=[(a-b)^2+(b-c)^2+(a-c)^2]/2

=(3^2+2^2+5^2)/2

=(9+4+25)/2

=19