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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 8: tìm Min: E=2x^2+9y^2-6xy-6x-12y+2004

Toán Lớp 8: tìm Min: E=2x^2+9y^2-6xy-6x-12y+2004

${E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l} x = 5\\ y = \dfrac{7}{3} \end{array} \right.$
$$\begin{array}{l} E = 2{x^2} + 9{y^2} – 6xy – 6x – 12y + 2004\\ = \left( {{x^2} – 6xy + 9{y^2}} \right) + \left( {4x – 12y} \right) + \left( {{x^2} – 10x + 25} \right) + 1979\\ = \left[ {{x^2} – 2.x.3y + {{\left( {3y} \right)}^2}} \right] + 4.\left( {x – 3y} \right) + \left( {{x^2} – 2.x.5 + {5^2}} \right) + 1979\\ = {\left( {x – 3y} \right)^2} + 4.\left( {x – 3y} \right) + {\left( {x – 5} \right)^2} + 1979\\ = \left[ {{{\left( {x – 3y} \right)}^2} + 4.\left( {x – 3y} \right) + 4} \right] + {\left( {x – 5} \right)^2} + 1975\\ = \left[ {{{\left( {x – 3y} \right)}^2} + 2.\left( {x – 3y} \right).2 + {2^2}} \right] + {\left( {x – 5} \right)^2} + 1975\\ = {\left[ {\left( {x – 3y} \right) + 2} \right]^2} + {\left( {x – 5} \right)^2} + 1975\\ = {\left( {x – 3y + 2} \right)^2} + {\left( {x – 5} \right)^2} + 1975\\ {\left( {x – 3y + 2} \right)^2} \ge 0,\,\,\,\,\forall x,y\\ {\left( {x – 5} \right)^2} \ge 0,\,\,\,\forall x\\ \Rightarrow {\left( {x – 3y + 2} \right)^2} + {\left( {x – 5} \right)^2} + 1975 \ge 1975,\,\,\,\,\forall x,y\\ \Rightarrow E \ge 1975,\,\,\,\forall x,y\\ \Rightarrow {E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l} {\left( {x – 3y + 2} \right)^2} = 0\\ {\left( {x – 5} \right)^2} = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x – 3y + 2 = 0\\ x = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 5\\ y = \dfrac{7}{3} \end{array} \right. \end{array}$$
Vậy $${E_{\min }} = 1975 \Leftrightarrow \left\{ \begin{array}{l} x = 5\\ y = \dfrac{7}{3} \end{array} \right.$$