Question

tại sao từ a^2 + b^2 + c^ - ab -bc - ca = 0 ta có thể suy ra \(\frac{1}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Answer 1

\(a^2+b^2+c^2-ac-ab-bc\)

\(=\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\)

\(=\frac{1}{2}\left(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right)\)

\(=\frac{1}{2}\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\)