Đáp án:
$\begin{array}{l}
a)Ox:y = 0\\
Khi:y = \left( {m - 1} \right).x + n//Ox\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 = 0\\
n\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 1\\
n\# 0
\end{array} \right.\\
Vậy\,m = 1;n\# 0\\
b) + Khi:m = - 2\\
Xet: - x + m + 2 = \left( {{m^2} - 2} \right).x + 1\\
\Leftrightarrow - x = 2.x + 1\\
\Leftrightarrow 3x = - 1\\
\Leftrightarrow x = - \dfrac{1}{3}\\
\Leftrightarrow y = - x + m + 2 = - \left( { - \dfrac{1}{3}} \right) - 2 + 2 = \dfrac{1}{3}\\
Vậy\,\left( d \right) \cap \left( {d'} \right) = \left( { - \dfrac{1}{3};\dfrac{1}{3}} \right)\,khi:m = - 2\\
+ \left( d \right)//\left( {d'} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
- 1 = {m^2} - 2\\
m + 2\# 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} = 1\\
m\# - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = 1/m = - 1\\
m\# - 1
\end{array} \right.\\
\Leftrightarrow m = 1\\
Vậy\,m = 1
\end{array}$
