Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
x \ge 0;x\# 9\\
y\# \dfrac{1}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{8}{{\sqrt x - 3}} + \dfrac{1}{{2y - 1}} = 5\\
\dfrac{4}{{\sqrt x - 3}} + \dfrac{1}{{2y - 1}} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{4}{{\sqrt x - 3}} = 2\\
\dfrac{1}{{2y - 1}} = \dfrac{4}{{\sqrt x - 3}} - 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x - 3 = 2\\
\dfrac{1}{{2y - 1}} = - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x = 5\\
2y - 1 = - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 25\left( {tmdk} \right)\\
y = 0\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {25;0} \right)\\
2)a)m = 0\\
\Leftrightarrow \left( d \right):y = - 3x + 4\\
Xet:{x^2} = - 3x + 4\\
\Leftrightarrow {x^2} + 3x - 4 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1;y = {x^2} = 1\\
x = - 4;y = {x^2} = 16
\end{array} \right.\\
Vậy\,\left( d \right) \cap \left( P \right):\left( {1;1} \right);\left( { - 4;16} \right)\\
b)Xet:{x^2} = \left( {m - 3} \right).x + 4\\
\Leftrightarrow {x^2} - \left( {m - 3} \right).x - 4 = 0\\
\Delta > 0\\
\Leftrightarrow {\left( {m - 3} \right)^2} + 16 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 3\\
{x_1}{x_2} = - 4
\end{array} \right.\\
Khi:{x_2} = \left| {{x_1}} \right|\\
\Leftrightarrow {x_2} = - {x_1}\left( {{x_1} < 0 < {x_2}} \right)\\
\Leftrightarrow - x_1^2 = - 4\\
\Leftrightarrow {x_1} = - 2\\
\Leftrightarrow {x_2} = 2\\
\Leftrightarrow 2 + \left( { - 2} \right) = m - 3\\
\Leftrightarrow m = 3\\
Vậy\,m = 3
\end{array}$
