{█((a-1)x+y=a@6x+ay=4)┤ tìm giá trị a để hệ phương trình có nghiệm thỏa x^2+y^2+xy<4

Giải thích các bước giải:

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