Với \(k\ne0\)
\(\Delta=\left(2k-1\right)^2-4k\left(k-2\right)=4k+1\ge0\Rightarrow k\ge-\frac{1}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\frac{-2k+1}{k}\\x_1x_2=\frac{k-2}{k}\end{matrix}\right.\)
\(x_1^2+x_2^2=2018\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=2018\)
\(\Leftrightarrow\frac{4k^2-4k+1}{k^2}-\frac{2k-4}{k}=2018\)
\(\Leftrightarrow4k^2-4k+1-2k^2+4k=2018k^2\)
\(\Leftrightarrow2016k^2=1\Rightarrow k=\pm\sqrt{\frac{1}{2016}}\)