\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
\(\Leftrightarrow a=b=c\)
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Ta có:
a2 + b2 + c2 = ab + bc + ac
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ac
=> 2a2 + 2b2 + 2c2 -2ab - 2bc - 2ac = 0
=> a2 + a2 + b2 + b2 + c2 + c2 -2ab - 2bc - 2ac = 0
=> ( a2 -2ab+b2 ) +( b2 - 2bc +c2 ) + ( a2 - 2ac + c2)=0
=> ( a-b)2 + ( b-c)2 + (a-c)2 =0
=> a-b =0 hoặc b-c = 0 hoặc a-c = 0
=> a = b = c ( đpcm)
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