Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - 2y = 4 - a\\
2x + y = 3a + 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2y = 4 - a\\
4x + 2y = 6a + 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5x = 10 + 5a\\
2x + y = 3a + 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = a + 2\\
y = 3a + 3 - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = a + 2\\
y = 3a + 3 - 2\left( {a + 2} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = a + 2\\
y = a - 1
\end{array} \right.
\end{array}$
Như vậy hệ phương trình có nghiệm là $\left( {x;y} \right) = \left( {a + 2;a - 1} \right)$
Khi đó:
$\begin{array}{l}
A = \dfrac{6}{{{x^2} + {y^2}}}\\
= \dfrac{6}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 1} \right)}^2}}}\\
= \dfrac{6}{{2{a^2} + 2a + 5}}\\
= \dfrac{{12}}{{4{a^2} + 4a + 10}}\\
= \dfrac{{12}}{{{{\left( {2a + 1} \right)}^2} + 9}}
\end{array}$
Mà ta có:
$\begin{array}{l}
{\left( {2a + 1} \right)^2} \ge 0,\forall a\\
\Rightarrow {\left( {2a + 1} \right)^2} + 9 \ge 9,\forall a\\
\Rightarrow \dfrac{{12}}{{{{\left( {2a + 1} \right)}^2} + 9}} \le \dfrac{{12}}{9} = \dfrac{4}{3}\\
\Rightarrow A \le \dfrac{4}{3}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow 2a + 1 = 0\\
\Leftrightarrow a = \dfrac{{ - 1}}{2}
\end{array}$
$ \Rightarrow x = \dfrac{3}{2};y = \dfrac{{ - 3}}{2}$
$ \Rightarrow MaxA = \dfrac{4}{3}$$ \Leftrightarrow x = \dfrac{3}{2};y = \dfrac{{ - 3}}{2}$
Vậy $a = \dfrac{{ - 1}}{2}$ thỏa mãn đề
