Đáp án:
$\begin{array}{l}
2)Dkxd:y > 0;y \ne 4\\
A = \dfrac{{4\sqrt y }}{{\sqrt y - 3}}\\
A = - 2\\
\Rightarrow \dfrac{{4\sqrt y }}{{\sqrt y - 3}} = - 2\\
\Rightarrow 4\sqrt y = - 2\sqrt y + 6\\
\Rightarrow 6\sqrt y = 6\\
\Rightarrow \sqrt y = 1\\
\Rightarrow y = 1\left( {tmdk} \right)\\
3)\left( d \right)//\left( {d'} \right)\\
\Rightarrow \left\{ \begin{array}{l}
a - 2 = 1\\
b - 1 \ne - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 3\\
b \ne - 2
\end{array} \right.\\
\Rightarrow y = x + b - 1\\
\left( {0; - 2} \right) \in \left( d \right)\\
\Rightarrow b - 1 = - 2\\
\Rightarrow b = - 1\\
\Rightarrow \left( d \right):y = x - 2\\
III)\\
2)\Delta ' > 0\\
\Rightarrow 1 - n + 3 > 0\\
\Rightarrow n < 4\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = n - 3
\end{array} \right.\\
x_1^2 - 2{x_1} + n - 3 = 0\\
\Rightarrow x_1^2 = 2{x_1} - n + 3\\
Do:x_1^2 - 2{x_2} + {x_1}{x_2} = - 12\\
\Rightarrow 2{x_1} - n + 3 - 2{x_2} + n - 3 = - 12\\
\Rightarrow {x_1} - {x_2} = - 6\\
\Rightarrow {x_2} - {x_1} = 6\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 36\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 36\\
\Rightarrow 4 - 4.\left( {n - 3} \right) = 36\\
\Rightarrow n = - 5\left( {tmdk} \right)
\end{array}$
