Đáp án:
b. \(Min = \dfrac{{3999}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = {x^2} + 10{y^2} + 6xy - 6y + 2020\\
= ({x^2} + 6xy + 9{y^2}) + ({y^2} - 6y + 9) + 2011\\
= {\left( {x + 3y} \right)^2} + {\left( {y - 3} \right)^2} + 2011\\
Do:\left\{ \begin{array}{l}
{\left( {x + 3y} \right)^2} \ge 0\forall x,y \in R\\
{\left( {y - 3} \right)^2} \ge 0\forall y \in R
\end{array} \right.\\
\to {\left( {x + 3y} \right)^2} + {\left( {y - 3} \right)^2} \ge 0\\
\to {\left( {x + 3y} \right)^2} + {\left( {y - 3} \right)^2} + 2011 \ge 2011\\
\to Min = 2011\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 3y = 0\\
y - 3 = 0
\end{array} \right. \to \left\{ \begin{array}{l}
y = 3\\
x = - 9
\end{array} \right.\\
B = {x^2} + 5{y^2} + 4xy - x - 11y + 2020\\
= {x^2} + {\left( {2y} \right)^2} + \dfrac{1}{4} + 2.x.2y - 2.x.\dfrac{1}{2} - 2.2y.\dfrac{1}{2} + {y^2} - 9y + \dfrac{{8079}}{4}\\
= {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {y^2} - 2.y.\dfrac{9}{2} + \dfrac{{81}}{4} + \dfrac{{3999}}{2}\\
= {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2}\\
Do:\left\{ \begin{array}{l}
{\left( {x + 2y - \dfrac{1}{2}} \right)^2} \ge 0\\
{\left( {y - \dfrac{9}{2}} \right)^2} \ge 0
\end{array} \right.\forall x,y \in R\\
\to {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} \ge 0\\
\to {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2} \ge \dfrac{{3999}}{2}\\
\to Min = \dfrac{{3999}}{2}\\
\Leftrightarrow \left\{ \begin{array}{l}
y - \dfrac{9}{2} = 0\\
x + 2y - \dfrac{1}{2} = 0
\end{array} \right. \to \left\{ \begin{array}{l}
y = \dfrac{9}{2}\\
x = - \dfrac{{17}}{2}
\end{array} \right.
\end{array}\)
