Đáp án:
\[{E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y = \dfrac{7}{3}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
E = 2{x^2} + 9{y^2} - 6xy - 6x - 12y + 2004\\
= \left( {{x^2} - 6xy + 9{y^2}} \right) + \left( {4x - 12y} \right) + \left( {{x^2} - 10x + 25} \right) + 1979\\
= \left[ {{x^2} - 2.x.3y + {{\left( {3y} \right)}^2}} \right] + 4.\left( {x - 3y} \right) + \left( {{x^2} - 2.x.5 + {5^2}} \right) + 1979\\
= {\left( {x - 3y} \right)^2} + 4.\left( {x - 3y} \right) + {\left( {x - 5} \right)^2} + 1979\\
= \left[ {{{\left( {x - 3y} \right)}^2} + 4.\left( {x - 3y} \right) + 4} \right] + {\left( {x - 5} \right)^2} + 1975\\
= \left[ {{{\left( {x - 3y} \right)}^2} + 2.\left( {x - 3y} \right).2 + {2^2}} \right] + {\left( {x - 5} \right)^2} + 1975\\
= {\left[ {\left( {x - 3y} \right) + 2} \right]^2} + {\left( {x - 5} \right)^2} + 1975\\
= {\left( {x - 3y + 2} \right)^2} + {\left( {x - 5} \right)^2} + 1975\\
{\left( {x - 3y + 2} \right)^2} \ge 0,\,\,\,\,\forall x,y\\
{\left( {x - 5} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {x - 3y + 2} \right)^2} + {\left( {x - 5} \right)^2} + 1975 \ge 1975,\,\,\,\,\forall x,y\\
\Rightarrow E \ge 1975,\,\,\,\forall x,y\\
\Rightarrow {E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3y + 2} \right)^2} = 0\\
{\left( {x - 5} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 3y + 2 = 0\\
x = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y = \dfrac{7}{3}
\end{array} \right.
\end{array}\)
Vậy \({E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y = \dfrac{7}{3}
\end{array} \right.\)
