Đáp án:
\[\left[ \begin{array}{l}
x = 1;y = - \frac{2}{9}\\
x = - \frac{6}{{17}};y = - \frac{{14}}{{17}}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
3{x^2} + 6xy - x + 3y = 0\,\,\,\,\,\left( 1 \right)\\
4x - 9y = 6\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \Leftrightarrow y = \frac{{4x - 6}}{9}\\
\left( 1 \right) \Leftrightarrow 3{x^2} + 6x.\frac{{4x - 6}}{9} - x + 3.\frac{{4x - 6}}{9} = 0\\
\Leftrightarrow 3{x^2} + \frac{2}{3}x.\left( {4x - 6} \right) - x + \frac{1}{3}\left( {4x - 6} \right) = 0\\
\Leftrightarrow 9{x^2} + 2x\left( {4x - 6} \right) - 3x + \left( {4x - 6} \right) = 0\\
\Leftrightarrow 9{x^2} + 8{x^2} - 12x - 3x + 4x - 6 = 0\\
\Leftrightarrow 17{x^2} - 11x - 6 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1 \Rightarrow y = - \frac{2}{9}\\
x = - \frac{6}{{17}} \Rightarrow y = - \frac{{14}}{{17}}
\end{array} \right.
\end{array}\)
