Ta có: x+y=-1
nên x=-1-y
Thay x=-1-y vào \(\dfrac{1}{x}-\dfrac{2}{y}=2\), ta được:
\(\dfrac{1}{-y-1}-\dfrac{2}{y}=2\)
\(\Leftrightarrow\dfrac{-1}{y+1}-\dfrac{2}{y}=2\)
\(\Leftrightarrow\dfrac{-y-2\left(y+1\right)}{y\left(y+1\right)}=\dfrac{2y\left(y+1\right)}{y\left(y+1\right)}\)
Suy ra: \(2y^2+2y+y+2y+2=0\)
\(\Leftrightarrow2y^2+4y+y+2=0\)
\(\Leftrightarrow\left(y+2\right)\left(2y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1-y=-1-\left(-2\right)=-1+2=1\\x=-1-y=-1-\left(-\dfrac{1}{2}\right)=-1+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-\dfrac{1}{2};-\dfrac{1}{2}\right)\right\}\)
