\(\Leftrightarrow\left\{{}\begin{matrix}k^2x-ky=2k\\x+ky=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(k^2+1\right)x=2k+1\\y=kx-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2k+1}{k^2+1}\\y=\dfrac{2k^2+k}{k^2+1}-2=\dfrac{-k}{k^2+1}\end{matrix}\right.\)
\(x+y=-1\Rightarrow\dfrac{2k+1}{k^2+1}+\dfrac{-k}{k^2+1}=-1\)
\(\Rightarrow k+1=-k^2-1\)
\(\Rightarrow k^2+k+2=0\) (vô nghiệm)
Không tồn tại k thỏa mãn yêu cầu