Đáp án:
$MinA = \dfrac{{24}}{{11}} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{{ - 19}}{{55}};\dfrac{1}{{11}}} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = 3{\left( {x - y} \right)^2} + 2{\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2}\\
= 3{x^2} - 6xy + 3{y^2} + 2{x^2} + 4x + 2 + {y^2} - 2y + 1\\
= 5{x^2} - 6xy + 4x + 4{y^2} - 2y + 3\\
= 5\left( {{x^2} - 2x.\dfrac{{3y - 2}}{5} + {{\left( {\dfrac{{3y - 2}}{5}} \right)}^2}} \right) - 5.{\left( {\dfrac{{3y - 2}}{5}} \right)^2} + 4{y^2} - 2y + 3\\
= 5{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} + \dfrac{{11}}{5}{y^2} + \dfrac{2}{5}y + \dfrac{{11}}{5}\\
= 5{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} + \dfrac{{11}}{5}\left( {{y^2} - 2y.\dfrac{1}{{11}} + \dfrac{1}{{121}}} \right) + \dfrac{{24}}{{11}}\\
= 5{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} + \dfrac{{11}}{5}{\left( {y - \dfrac{1}{{11}}} \right)^2} + \dfrac{{24}}{{11}}
\end{array}$
Lại có:
$\begin{array}{l}
{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2},{\left( {y - \dfrac{1}{{11}}} \right)^2} \ge 0,\forall x,y\\
\Rightarrow 5{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} + \dfrac{{11}}{5}{\left( {y - \dfrac{1}{{11}}} \right)^2} \ge 0,\forall x,y\\
\Rightarrow 5{\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} + \dfrac{{11}}{5}{\left( {y - \dfrac{1}{{11}}} \right)^2} + \dfrac{{24}}{{11}} \ge \dfrac{{24}}{{11}},\forall x,y\\
\Rightarrow A \ge \dfrac{{24}}{{11}}\\
\Rightarrow MinA = \dfrac{{24}}{{11}}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow {\left( {x - \dfrac{{3y - 2}}{5}} \right)^2} = {\left( {y - \dfrac{1}{{11}}} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x - \dfrac{{3y - 2}}{5} = 0\\
y - \dfrac{1}{{11}} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = \dfrac{1}{{11}}\\
x = \dfrac{{ - 19}}{{55}}
\end{array} \right.
\end{array}$
Vậy $MinA = \dfrac{{24}}{{11}} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{{ - 19}}{{55}};\dfrac{1}{{11}}} \right)$
