tìm GTNN,GTLN E=2X^2+5Y^2-4XY-4X+2Y+3

$\begin{array}{l} E = 2{x^2} + 5{y^2} - 4xy - 4x + 2y + 3\\ = {x^2} - 4xy + 4{y^2} + {x^2} - 4x + 4 + {y^2} + 2y + 1 - 2\\ = {\left( {x - 2y} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} - 2\\ Do\,\,{\left( {x - 2y} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\\ \Rightarrow E \ge - 2\\ GTNN\,cua\,E\,la\, - 2 \Leftrightarrow \left\{ \begin{array}{l} x = 2y\\ x = 2\\ y = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 1 \end{array} \right. \end{array}$

Đáp án:

Giải thích các bước giải: $\begin{array}{l} E = 2{x^2} + 5{y^2} - 4xy - 4x + 2y + 3\\ = {x^2} - 4xy + 4{y^2} + {x^2} - 4x + 4 + {y^2} + 2y + 1 - 2\\ = {\left( {x - 2y} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} - 2\\ Do\,\,{\left( {x - 2y} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\\ \Rightarrow E \ge - 2\\ GTNN\,cua\,E\,la\, - 2 \Leftrightarrow \left\{ \begin{array}{l} x = 2y\\ x = 2\\ y = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 1 \end{array} \right. \end{array}$