Question

[manh nguyenvan]

a^2+b^2+c^2 -ab -bc -ca =0.C/m a=b=c

Answer 1

[missing you =]

\(< =>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(< =>a^2-2ab+b^2+a^2-2ca+c^2+b^2-2bc+c^2=0\)

\(< =>\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

có \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{matrix}\right.\) dấu"=" xảy ra<=>a=b=c

Answer 2

[Nguyễn Lê Phước Thịnh]

Ta có: \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Suy ra: a=b=c