Đáp án:
\[{A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{3}{2}\\
y = 3
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = - \left| {2x + 3} \right| - \dfrac{1}{2}\left| {3 - y} \right| + \dfrac{3}{5}\\
= \dfrac{3}{5} - \left( {\left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 - y} \right|} \right)\\
\left. \begin{array}{l}
\left| {2x + 3} \right| \ge 0,\,\,\,\forall x\\
\left| {3 - y} \right| \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow \left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 - y} \right| \ge 0,\,\,\,\,\forall x,y\\
\Rightarrow \dfrac{3}{5} - \left( {\left| {2x + 3} \right| + \dfrac{1}{2}\left| {3 - y} \right|} \right) \le \dfrac{3}{5},\,\,\,\forall x,y\\
\Rightarrow {A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
\left| {2x + 3} \right| = 0\\
\left| {3 - y} \right| = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{3}{2}\\
y = 3
\end{array} \right.
\end{array}\)
Vậy \({A_{\max }} = \dfrac{3}{5} \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{3}{2}\\
y = 3
\end{array} \right.\)
