Ta có: \(\left\{{}\begin{matrix}x+y=-1\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{1}{-y-1}-\dfrac{2}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{-1}{y+1}-\dfrac{2}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{-y-2\left(y+1\right)}{y\left(y+1\right)}=\dfrac{2y^2+2y}{y\left(y+1\right)}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\-y-2y-2=2y^2+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-1\\2y^2+5y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y-1\\\left(2y+1\right)\left(y+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}-1=-\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\left(-2\right)-1=2-1=1\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\left\{\left(-\dfrac{1}{2};-\dfrac{1}{2}\right);\left(1;-2\right)\right\}\)
ĐK : x ≠ 0 ; y ≠ 0
\(\left\{{}\begin{matrix}x+y=-1\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}x=-1-y\\\dfrac{1}{-1-y}-\dfrac{2}{y}=2\end{matrix}\right.\)\(\text{⇔}\left\{{}\begin{matrix}x=-1-y\\y-2\left(-1-y\right)=2y\left(-1-y\right)\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}x=-1-y\\y+2+2y=-2y-2y^2\end{matrix}\right.\)
\(\text{⇔}\left\{{}\begin{matrix}x=-1-y\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Với \(y=-\dfrac{1}{2}\text{⇔}x=-\dfrac{1}{2}\)
Với \(y=-2\text{⇔}x=1\)
Vậy.....
