\[\begin{array}{l}
a){x^2} - 2x - 5 = 0\\
\Leftrightarrow {x^2} - 2x + 1 - 6 = 0\\
\Leftrightarrow {(x - 1)^2} = 6\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = \sqrt 6 \\
x - 1 = - \sqrt 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1 + \sqrt 6 \\
x = 1 - \sqrt 6
\end{array} \right.\\
Vay:S = \{ 1 + \sqrt 6 ;1 - \sqrt 6 \} \\
b){x^4} - 4{x^2} + 3 = 0\\
\Leftrightarrow {x^4} - 4{x^2} + 4 - 1 = 0\\
\Leftrightarrow {({x^2} - 2)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2 = 1\\
{x^2} - 2 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} = 3\\
{x^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3 \\
x = 1\\
x = - 1
\end{array} \right.\\
Vay:S = \{ \sqrt 3 ; - \sqrt 3 ;1; - 1\} \\
c){x^2} - 3x - 1 = 0\\
\Leftrightarrow {x^2} - 3x + \frac{9}{4} - \frac{{13}}{4} = 0\\
\Leftrightarrow {(x - \frac{3}{2})^2} = \frac{{13}}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \frac{3}{2} = \frac{{\sqrt {13} }}{2}\\
x - \frac{3}{2} = \frac{{ - \sqrt {13} }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{3 + \sqrt {13} }}{2}\\
x = \frac{{3 - \sqrt {13} }}{2}
\end{array} \right.\\
vay:S = \{ \frac{{3 + \sqrt {13} }}{2};\frac{{3 - \sqrt {13} }}{2}\} \\
d){x^2} + xy + {y^2} = 0\\
\Leftrightarrow {x^2} + xy + \frac{{{y^2}}}{4} + \frac{{3{y^3}}}{4} = 0\\
\Leftrightarrow {(x + \frac{y}{2})^2} = - \frac{{3{y^2}}}{4}\\
do:{(x + \frac{y}{2})^2} \ge 0\\
- \frac{{3{y^2}}}{4} \le 0
\end{array}\]

Dấu = xảy ra \[ \Leftrightarrow \left\{ \begin{array}{l}
x + \frac{y}{2} = 0\\
- \frac{{3{y^2}}}{4} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 0
\end{array} \right.\]

vậy nghiệm pt là (x;y)=(0;0)