Đáp án: $Min A=- \dfrac{{19}}{4}\Leftrightarrow (x;y)=\left(\dfrac{5}{2};\dfrac{-7}{2}\right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = 2{{\rm{x}}^2} + {y^2} + 2{\rm{x}}y - 3{\rm{x}} + 5 + 2y\\
\Leftrightarrow 2{\rm{A}} = 4{{\rm{x}}^2} + 2{y^2} + 4{\rm{x}}y - 6{\rm{x}} + 10 + 4y\\
\Leftrightarrow 2{\rm{A}} = {(2{\rm{x}})^2} + 2.2{\rm{x}}\left( {y - \dfrac{3}{2}} \right) + {\left( {y - \dfrac{3}{2}} \right)^2} + {y^2} + 7y + \dfrac{{11}}{4}\\
\Leftrightarrow 2{\rm{A}} = {\left( {2{\rm{x}} + y - \dfrac{3}{2}} \right)^2} + {\left( {y + \dfrac{7}{2}} \right)^2} - \dfrac{{19}}{2}
\end{array}$
$\to Min 2A=- \dfrac{{19}}{2}\to Min A=- \dfrac{{19}}{4}$
Dấu bằng xảy ra khi và chỉ khi: $\left\{ \begin{array}{l}
2{\rm{x}} + y - \dfrac{3}{2} = 0\\
y + \dfrac{7}{2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{2}\\
y = \dfrac{{ - 7}}{2}
\end{array} \right.$
Vậy $Min A=- \dfrac{{19}}{4}\Leftrightarrow (x;y)=\left(\dfrac{5}{2};\dfrac{-7}{2}\right)$
