a) \(a^2+b^2+c^2=ab+bc+ca\)
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
=> \(a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0\)
<=> (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 -2bc + c2) = 0
<=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\) (1)
Mà \(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\) (2)
Từ (1); (2) =>
+ \(\left(a-b\right)^2=0\Leftrightarrow a=b\)
+ \(\left(a-c\right)^2=0\Leftrightarrow a=c\)
+ \(\left(b-c\right)^2=0\Leftrightarrow b=c\)
=> a = b = c => đpcm
2 Câu dưới tương tự bài a bn nhé