Tham khảo tại link sau : http://olm.vn/hoi-dap/question/721476.html
a2+b2+c2=ab+bc+ac
2a2+2b2+2c2=2ab+2bc+2ac
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a-b)2+(a-c)2+(b-c)2=0
=> a=b=c
k co mình cái
\(a^2+b^2+c^2=ac+bc+ab\)
\(a^2+b^2+c^2-ab-ac-bc=0\)
\(2a^2+2b^2+2c^2-2ac-2bc-2ab=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
suy ra a=b=c