Đáp án:
Min=2012
Giải thích các bước giải:
\(\begin{array}{l}
A = {x^2} + {\left( {\dfrac{1}{2}y} \right)^2} + {\left( { - 1} \right)^2} + 2.x.\dfrac{1}{2}y - 2.x - 2.\dfrac{1}{2}y + \dfrac{3}{4}{y^2} - 2.\dfrac{{\sqrt 3 }}{2}y.\sqrt 3 + 3 + 2012\\
= {\left( {x + \dfrac{1}{2}y - 1} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}y - \sqrt 3 } \right)^2} + 2012\\
Do:\left\{ \begin{array}{l}
{\left( {x + \dfrac{1}{2}y - 1} \right)^2} \ge 0\forall x;y\\
{\left( {\dfrac{{\sqrt 3 }}{2}y - \sqrt 3 } \right)^2} \ge 0\forall x
\end{array} \right.\\
\to {\left( {x + \dfrac{1}{2}y - 1} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}y - \sqrt 3 } \right)^2} \ge 0\\
\to {\left( {x + \dfrac{1}{2}y - 1} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}y - \sqrt 3 } \right)^2} + 2012 \ge 2012\\
\to Min = 2012\\
\Leftrightarrow \left\{ \begin{array}{l}
x + \dfrac{1}{2}y - 1 = 0\\
\dfrac{{\sqrt 3 }}{2}y - \sqrt 3 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 2\\
x = 0
\end{array} \right.
\end{array}\)
