Vì \(\dfrac{-2}{x}=\dfrac{y}{6}\)
\(\Rightarrow xy=-2.6\)
\(\Rightarrow xy=-12\)
Mà x<0<y
\(\Rightarrow\) x \(\in\) {-1;-2;-3;-4;-6;-12}
\(\Rightarrow\) y \(\in\) {12; 6; 4; 3; 2; 1}
Vậy (x,y)=(-1,12);(-2;6);(-3;4);(-4;3);(-6;2);(-12;1) thỏa mãn bài ra
\(\dfrac{-2}{x}=\dfrac{y}{6}\)
\(\Rightarrow xy=6.-2\)
\(\Rightarrow xy=-12\)
\(\Rightarrow xy\inƯ\left(-12\right)\)
\(Ư\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-12\\x=-1\\y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-6\\x=-2\\y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-4\\x=-3\\y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-3\\x=-4\\y=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-2\\x=-6\\y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=12\\y=-1\\x=-12\\y=1\end{matrix}\right.\)
