\(\left\{{}\begin{matrix}x+y=-1\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\) ( Điều kiện : \(x\ne0;y\ne0\) )
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{1}{-1-y}-\dfrac{2}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{y}{y\left(-1-y\right)}-\dfrac{2\left(-1-y\right)}{y\left(-1-y\right)}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{y}{y\left(-1-y\right)}-\dfrac{-2-2y}{y\left(-1-y\right)}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{y+2+2y}{y\left(-1-y\right)}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{3y+2}{-y-y^2}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\2\left(-y-y^2\right)=3y+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\-2y-2y^2-3y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\-2y^2-5y-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=-2\end{matrix}\right.\end{matrix}\right.\)
* Nếu \(y=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{2}\) ( TMĐK )
* Nếu \(y=-2\Rightarrow x=1\) ( TMĐK )
