Đáp án:
\[\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 1\\
y = - 7
\end{array} \right.
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3{x^2} + 3xy - 17 = 7x - 2y\\
\Leftrightarrow 3x\left( {x + y} \right) - 17 = 9x - 2\left( {x + y} \right)\\
\Leftrightarrow \left( {x + y} \right)\left( {3x + 2} \right) = 17 + 9x\\
\Leftrightarrow \left( {x + y} \right)\left( {3x + 2} \right) = 3\left( {3x + 2} \right) + 11\\
\Leftrightarrow \left( {3x + 2} \right)\left( {x + y - 3} \right) = 11\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x + 2 = 1\\
x + y - 3 = 11
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 2 = 11\\
x + y - 3 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 2 = - 1\\
x + y - 3 = - 11
\end{array} \right.\\
\left\{ \begin{array}{l}
3x + 2 = - 11\\
x + y - 3 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = - \frac{1}{3}\\
y = \frac{{43}}{3}
\end{array} \right.\left( L \right)\\
\left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\left( {t.m} \right)\\
\left\{ \begin{array}{l}
x = - 1\\
y = - 7
\end{array} \right.\left( {t/m} \right)\\
\left\{ \begin{array}{l}
x = - \frac{{13}}{3}\\
y = \frac{{19}}{3}
\end{array} \right.\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 1\\
y = - 7
\end{array} \right.
\end{array} \right.
\end{array}\)
