Câu 6:
\(x-3=y\left(x+2\right)\\ \Leftrightarrow x+2-5-y\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(1-y\right)=5=5.1=\left(-5\right)\left(-1\right)\)
Với \(\left\{{}\begin{matrix}x+2=1\\1-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x+2=5\\1-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x+2=-5\\1-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=2\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x+2=-1\\1-y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=6\end{matrix}\right.\)
Vậy các cặp nghiệm \(\left(x;y\right)\) là \(\left(-1;-4\right);\left(3;0\right);\left(-7;2\right);\left(-3;6\right)\)
