Đáp án:
\(a=b=c=2.\)
Giải thích các bước giải:
\(\begin{array}{l}
{\left( {a + b + c} \right)^2} + 12 = 4\left( {a + b + c} \right) + 2\left( {ab + ac + bc} \right)\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca + 12 = 4\left( {a + b + c} \right) + 2ab + 2bc + 2ca\\
\Leftrightarrow {a^2} + {b^2} + {c^2} - 4a - 4b - 4c + 12 = 0\\
\Leftrightarrow \left( {{a^2} - 4a + 4} \right) + \left( {{b^2} - 4b + 4} \right) + \left( {{c^2} - 4c + 4} \right) = 0\\
\Leftrightarrow {\left( {a - 2} \right)^2} + {\left( {b - 2} \right)^2} + {\left( {c - 2} \right)^2} = 0\,\,\left( * \right)\\
Vi\,\,\,\left\{ \begin{array}{l}
{\left( {a - 2} \right)^2} \ge 0\\
{\left( {b - 2} \right)^2} \ge 0\\
{\left( {c - 2} \right)^2} \ge 0
\end{array} \right.\,\,\,\,\forall a,\,\,b,\,\,c\\
\left( * \right)\,\,\,co\,\,\,nghiem\,\, \Leftrightarrow \left\{ \begin{array}{l}
a - 2 = 0\\
b - 2 = 0\\
c - 2 = 0
\end{array} \right. \Leftrightarrow a = b = c = 2.
\end{array}\)
