Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {x^2} + 10{y^2} + 6xy - 6y + 2020\\
= \left( {{x^2} + 6xy + 9{y^2}} \right) + \left( {{y^2} - 6y + 9} \right) + 2011\\
= \left[ {{x^2} + 2x.3y + {{\left( {3y} \right)}^2}} \right] + \left[ {{y^2} - 2.y.3 + {3^2}} \right] + 2011\\
= {\left( {x + 3y} \right)^2} + {\left( {y - 3} \right)^2} + 2011\\
\left. \begin{array}{l}
{\left( {x + 3y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {y - 3} \right)^2} \ge 0,\,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {x + 3y} \right)^2} + {\left( {y - 3} \right)^2} + 2011 \ge 2011,\,\,\,\forall x,y\\
\Rightarrow {A_{\min }} = 2011 \Leftrightarrow \left\{ \begin{array}{l}
x + 3y = 0\\
y - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 9\\
y = 3
\end{array} \right.\\
b,\\
B = {x^2} + 5{y^2} + 4xy - x - 11y + 2020\\
= \left( {{x^2} + 4xy + 4{y^2}} \right) - \left( {x + 2y} \right) + \dfrac{1}{4} + \left( {{y^2} - 9y + \dfrac{{81}}{4}} \right) + \dfrac{{3999}}{2}\\
= {\left( {x + 2y} \right)^2} - \left( {x + 2y} \right) + \dfrac{1}{4} + \left[ {{y^2} - 2.y.\dfrac{9}{2} + {{\left( {\dfrac{9}{2}} \right)}^2}} \right] + \dfrac{{3999}}{2}\\
= \left[ {{{\left( {x + 2y} \right)}^2} - 2.\left( {x + 2y} \right).\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2}\\
= {\left( {x + 2y - \dfrac{1}{2}} \right)^2} + {\left( {y - \dfrac{9}{2}} \right)^2} + \dfrac{{3999}}{2} \ge \dfrac{{3999}}{2},\,\,\,\forall x,y\\
\Rightarrow {A_{\min }} = \dfrac{{3999}}{2} \Leftrightarrow \left\{ \begin{array}{l}
x + 2y - \dfrac{1}{2} = 0\\
y - \dfrac{9}{2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{{17}}{2}\\
y = \dfrac{9}{2}
\end{array} \right.
\end{array}\)
