Toán Chứng minh 4x^2+y^2 +z^2 +14 >= 12x-2y-4z 04/10/2021 By Autumn Chứng minh 4x^2+y^2 +z^2 +14 >= 12x-2y-4z
\[\begin{array}{l} 4{x^2} + {y^2} + {z^2} + 14 \ge 12x – 2y – 4z\\ \Leftrightarrow 4{x^2} – 12x + {y^2} – 2y + {z^2} – 4z + 14 \ge 0\\ \Leftrightarrow 4{x^2} – 12x + 9 + {y^2} – 2y + 1 + {z^2} – 4z + 4 \ge 0\\ \Leftrightarrow {\left( {2x – 3} \right)^2} + {\left( {y – 1} \right)^2} + {\left( {z – 2} \right)^2} \ge 0. \end{array}\] Trả lời
Đáp án:
(2x−3)2+(y−1)2+(z−2)2≥0.
\[\begin{array}{l}
4{x^2} + {y^2} + {z^2} + 14 \ge 12x – 2y – 4z\\
\Leftrightarrow 4{x^2} – 12x + {y^2} – 2y + {z^2} – 4z + 14 \ge 0\\
\Leftrightarrow 4{x^2} – 12x + 9 + {y^2} – 2y + 1 + {z^2} – 4z + 4 \ge 0\\
\Leftrightarrow {\left( {2x – 3} \right)^2} + {\left( {y – 1} \right)^2} + {\left( {z – 2} \right)^2} \ge 0.
\end{array}\]