Đáp án:
$\begin{array}{l}
{x^2} - 2xy + 5{y^2} = y + 1\\
\Leftrightarrow {x^2} - 2y.x + \left( {5{y^2} - y - 1} \right) = 0\\
\Rightarrow \left\{ \begin{array}{l}
a = 1\\
b = - 2y\\
c = 5{y^2} - y - 1
\end{array} \right.\\
\Rightarrow \Delta ' = {y^2} - 5{y^2} + y + 1 = - 4{y^2} + y + 1
\end{array}$