Đáp án:
$\begin{array}{l}
2)\left\{ \begin{array}{l}
x + 3y = 3\\
- x + 2y = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5y = 10\\
x = 3 - 3y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 2\\
x = - 3
\end{array} \right.\\
Vậy\,x = - 3;y = 2\\
3)a){x^2} - 2x - 3 = 0\\
\Leftrightarrow {x^2} - 3x + x - 3 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
Vậy\,x = - 1;x = 3\\
b)\Delta ' > 0\\
\Leftrightarrow 1 - m > 0\\
\Leftrightarrow m < 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m
\end{array} \right.\\
Khi:x_1^2 + x_2^2 = 8\\
\Leftrightarrow x_1^2 + 2{x_1}{x_2} + x_2^2 - 2{x_1}{x_2} = 8\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 8\\
\Leftrightarrow {2^2} - 2m = 8\\
\Leftrightarrow 2m = - 4\\
\Leftrightarrow m = - 2\left( {tmdk} \right)\\
Vậy\,m = - 2
\end{array}$
