Đáp án:
1) m=1
Giải thích các bước giải:
\(\begin{array}{l}
a)Do:\left( d \right)//\left( {d'} \right)\\
\to \left\{ \begin{array}{l}
- 2 = {m^2} - 3\\
m + 3 \ne 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} = 1\\
m \ne - 1
\end{array} \right.\\
\to m = 1\\
2)Xet:\Delta ' \ge 0\\
\to 1 - m + 3 \ge 0\\
\to 4 \ge m\\
Có:{x_1}\left( {2 - {x_1}^2} \right) + {x_2}\left( {2 - {x_2}^2} \right) = {\left( {{x_1}{x_2}} \right)^2} - 31\\
\to - {x_1}^3 + 2{x_1} + 2{x_2} - {x_2}^3 = {\left( {{x_1}{x_2}} \right)^2} - 31\\
\to - \left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 - {x_1}{x_2} + {x_2}^2} \right) + 2\left( {{x_1} + {x_2}} \right) = {\left( {{x_1}{x_2}} \right)^2} - 31\\
\to - \left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 3{x_1}{x_2}} \right) + 2\left( {{x_1} + {x_2}} \right) = {\left( {{x_1}{x_2}} \right)^2} - 31\\
\to - \left( {{x_1} + {x_2}} \right)\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 3{x_1}{x_2}} \right] + 2\left( {{x_1} + {x_2}} \right) = {\left( {{x_1}{x_2}} \right)^2} - 31\\
\to - 2\left( {4 - 3\left( {m - 3} \right)} \right) + 2.2 = {\left( {m - 3} \right)^2} - 31\\
\to {m^2} - 6m + 9 - 31 = - 2\left( {4 - 3m + 9} \right) + 4\\
\to {m^2} - 12m - 8 = 0\\
\to \left[ \begin{array}{l}
m = 6 + 2\sqrt {11} \left( l \right)\\
m = 6 - 2\sqrt {11}
\end{array} \right.
\end{array}\)
