Giải thích các bước giải:
\(\begin{array}{l}
D = 2x + 12y + 6z - {x^2} - 4{y^2} - {z^2} - 18\\
= - \left( {{x^2} - 2x + 1} \right) - \left( {4{y^2} - 12y + 9} \right) - \left( {{z^2} - 6z + 9} \right) + 1\\
= - {\left( {x - 1} \right)^2} - {\left( {2y - 3} \right)^2} - {\left( {z - 3} \right)^2} + 1\\
= 1 - \left[ {{{\left( {x - 1} \right)}^2} + {{\left( {2y - 3} \right)}^2} + {{\left( {z - 3} \right)}^2}} \right]\\
{\left( {x - 1} \right)^2} + {\left( {2y - 3} \right)^2} + {\left( {z - 3} \right)^2} \ge 0,\,\,\forall x,y,z\\
\Rightarrow 1 - \left[ {{{\left( {x - 1} \right)}^2} + {{\left( {2y - 3} \right)}^2} + {{\left( {z - 3} \right)}^2}} \right] \le 1,\,\,\,\forall x,y,z\\
\Rightarrow D \le 1,\,\,\,\forall x,y,z\\
\Rightarrow {D_{\max }} = 1 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} = 0\\
{\left( {2y - 3} \right)^2} = 0\\
{\left( {z - 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = \frac{3}{2}\\
z = 3
\end{array} \right.
\end{array}\)
Em xem lại đề ý C nhé.
