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