Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {a - 1} \right)x + y = a\\
6x + ay = 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = a - \left( {a - 1} \right)x\\
6x + a\left( {a - \left( {a - 1} \right)} \right)x = 4
\end{array} \right.\\
\Rightarrow \left( {{a^2} - a - 6} \right)x = {a^2} - 4\\
\Rightarrow x = \dfrac{{{a^2} - 4}}{{{a^2} - a - 6}} = \dfrac{{\left( {a - 2} \right)\left( {a + 2} \right)}}{{\left( {a + 2} \right)\left( {a - 3} \right)}} = \dfrac{{a - 2}}{{a - 3}}\left( {a \ne - 2;a \ne 3} \right)\\
\Rightarrow y = a - \left( {a - 1} \right)x = a - \left( {a - 1} \right).\dfrac{{a - 2}}{{a - 3}} = \dfrac{{{a^2} - 3a - \left( {{a^2} - 3a + 2} \right)}}{{a - 3}} = \dfrac{{ - 2}}{{a - 3}}\\
{x^2} + {y^2} + xy < 4\\
\Leftrightarrow \dfrac{{{{\left( {a - 2} \right)}^2}}}{{{{\left( {a - 3} \right)}^2}}} + \dfrac{4}{{{{\left( {a - 3} \right)}^2}}} - \dfrac{{2a - 4}}{{{{\left( {a - 3} \right)}^2}}} < 4\\
\Leftrightarrow \dfrac{{{a^2} - 6a + 12}}{{{{\left( {a - 3} \right)}^2}}} - 4 < 0\\
\Leftrightarrow \dfrac{{{a^2} - 6a + 12 - 4{a^2} + 24a - 36}}{{{{\left( {a - 3} \right)}^2}}} < 0\\
\Rightarrow - 3{a^2} + 18a - 24 < 0\\
\Leftrightarrow 3{a^2} - 18a + 24 > 0\\
\Leftrightarrow 3\left( {a - 2} \right)\left( {a - 4} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a - 2 > 0\\
a - 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
a - 2 < 0\\
a - 4 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a > 4\\
a < 2
\end{array} \right.
\end{array}\)
