1) \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(VT\ge0\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
Vậy tam giác đã cho đều
2) \(P=a^3+b^3+a^2c+b^2c-abc\)
\(P=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2+b^2-ab\right)\)
\(P=\left(a^2+b^2-ab\right)\left(a+b+c\right)\)
\(P=0\)