Đáp án:
\(Min = \dfrac{{3999}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
B = {x^2} + 5{y^2} + 4xy - x - 11y + 2020\\
= {x^2} + {\left( {2y} \right)^2} + \dfrac{1}{4} + 2.x.2y - 2.x.\dfrac{1}{2} - 2.2y.\dfrac{1}{2} + {y^2} - 9y + \dfrac{{8079}}{4}\\
= {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {y^2} - 2.y.\dfrac{9}{2} + \dfrac{{81}}{4} + \dfrac{{3999}}{2}\\
= {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2}\\
Do:\left\{ {\begin{array}{*{20}{l}}
{{{\left( {x + 2y - \dfrac{1}{2}} \right)}^2} \ge 0}\\
{{{\left( {y - \dfrac{9}{2}} \right)}^2} \ge 0}
\end{array}} \right.\forall x,y \in R\\
\to {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} \ge 0\\
\to {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2} \ge \dfrac{{3999}}{2}\\
\to Min = \dfrac{{3999}}{2}\\
\Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{y - \dfrac{9}{2} = 0}\\
{x + 2y - \dfrac{1}{2} = 0}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{l}}
{y = \dfrac{9}{2}}\\
{x = {\rm{ \;}} - \dfrac{{17}}{2}}
\end{array}} \right.
\end{array}\)
