Đáp án:
a. m=-2
Giải thích các bước giải:
\(\begin{array}{l}
a.Do:\left( D \right)//\left( {{d_1}} \right)\\
\to \left\{ \begin{array}{l}
{m^2} - 2 = 2\\
m - 1 \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} = 4\\
m \ne 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 2\left( l \right)\\
m = - 2
\end{array} \right.
\end{array}\)
b. Có:
\(\begin{array}{l}
\left( {{d_2}} \right):y = 2mx - m + 3 + x\\
\to y = \left( {2m + 1} \right)x + 3 - m
\end{array}\)
Do (D) cắt (d2)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
{m^2} - 2 \ne 2m + 1\\
m - 1 \ne 3 - m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \left\{ { - 1;3} \right\}\\
2m \ne 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \left\{ { - 1;3} \right\}\\
m \ne 2
\end{array} \right.
\end{array}\)
