Đáp án:
$\begin{array}{l}
2)1)\\
x = 2 + \sqrt {2022} \\
\Leftrightarrow {\left( {2 + \sqrt {2022} } \right)^2} - m\left( {2 + \sqrt {2022} } \right) - 2m - 4 = 0\\
\Leftrightarrow - \left( {4 + \sqrt {2022} } \right).m + \left( {4\sqrt {2022} + 2026 - 4} \right) = 0\\
\Leftrightarrow \left( {4 + \sqrt {2022} } \right).m = \left( {4\sqrt {2022} + 2022} \right)\\
\Leftrightarrow \left( {4 + \sqrt {2022} } \right).m = \sqrt {2022} .\left( {4 + \sqrt {2022} } \right)\\
\Leftrightarrow m = \sqrt {2022} \\
Vậy\,m = \sqrt {2022} \\
2)\Delta > 0\\
\Leftrightarrow {m^2} - 4\left( { - 2m - 4} \right) > 0\\
\Leftrightarrow {m^2} + 8m + 16 > 0\\
\Leftrightarrow {\left( {m + 4} \right)^2} > 0\\
\Leftrightarrow m\# - 4\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = - 2m - 4
\end{array} \right.\\
x_2^2 - m{x_2} - 2m - 4 = 0\\
\Leftrightarrow x_2^2 = m{x_2} + 2m + 4\\
Khi:x_2^2 + 2{x_1}{x_2} + m{x_1} = 20\\
\Leftrightarrow m{x_2} + 2m + 4 + 2{x_1}{x_2} + m{x_1} = 20\\
\Leftrightarrow m\left( {{x_1} + {x_2}} \right) + 2{x_1}{x_2} + 2m - 16 = 0\\
\Leftrightarrow m.m + 2.\left( { - 2m - 4} \right) + 2m - 16 = 0\\
\Leftrightarrow {m^2} - 2m - 24 = 0\\
\Leftrightarrow \left( {m - 6} \right)\left( {m + 4} \right) = 0\\
\Leftrightarrow m = 6\left( {tm} \right);m = - 4\left( {ktm} \right)\\
Vậy\,m = 6\\
3)Dkxd:\left\{ \begin{array}{l}
x \ge 2y\\
y \ge 0
\end{array} \right. \Leftrightarrow x \ge 2y \ge 0\\
\left\{ \begin{array}{l}
{x^2} + 2xy = 15{y^2}\\
\sqrt {x - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2xy - 15{y^2} = 0\\
\sqrt {x - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 5xy + 3xy - 15{y^2} = 0\\
\sqrt {x - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 5y} \right)\left( {x + 3y} \right) = 0\\
\sqrt {x - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5y\left( {do:x + 3y > 0} \right)\\
\sqrt {x - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5y\\
\sqrt {5y - 2y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5y\\
\sqrt {3y} + \sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5y\\
\left( {\sqrt 3 + 1} \right).\sqrt y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5y\\
y = {\left( {\dfrac{2}{{\sqrt 3 + 1}}} \right)^2} = {\left( {\sqrt 3 - 1} \right)^2} = 4 - 2\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 20 - 10\sqrt 3 \\
y = 4 - 2\sqrt 3
\end{array} \right.\\
Vậy\,x = 20 - 10\sqrt 2 ;y = 4 - 2\sqrt 3
\end{array}$
