2 câu trả lời
\[\begin{array}{l} 4{x^2} + {y^2} + {z^2} + 14 \ge 12x - 2y - 4z\\ \Leftrightarrow 4{x^2} - 12x + {y^2} - 2y + {z^2} - 4z + 14 \ge 0\\ \Leftrightarrow 4{x^2} - 12x + 9 + {y^2} - 2y + 1 + {z^2} - 4z + 4 \ge 0\\ \Leftrightarrow {\left( {2x - 3} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {z - 2} \right)^2} \ge 0. \end{array}\]