Đáp án:
a) Xét pt hoành độ giao điểm:
$\begin{array}{l}
\left( {m + 1} \right)x + 1 = 2x + 2\\
\Leftrightarrow \left( {m + 1 - 2} \right).x = 1\\
\Leftrightarrow \left( {m - 1} \right).x = 1\\
+ Khi:m = 1 \Leftrightarrow 0.x = 1\left( {ktm} \right)\\
+ Khi;m\# 1 \Leftrightarrow x = \dfrac{1}{{m - 1}}\\
\Leftrightarrow y = 2x + 2 = 2.\dfrac{1}{{m - 1}} + 2 = \dfrac{{2m}}{{m - 1}}\\
\Leftrightarrow \left( {{d_1}} \right) \cap \left( {{d_2}} \right) = \left( {\dfrac{1}{{m - 1}};\dfrac{{2m}}{{m - 1}}} \right)\\
b)\left( {{d_1}} \right) \cap \left( {{d_2}} \right) = \left( {{x_0};{y_o}} \right)\\
{x_0}.{y_0} > 0\\
\Leftrightarrow \dfrac{1}{{m - 1}}.\dfrac{{2m}}{{m - 1}} > 0\\
\Leftrightarrow \dfrac{{2m}}{{{{\left( {m - 1} \right)}^2}}} > 0\\
\Leftrightarrow m > 0\\
Vậy\,m > 0;m\# 1
\end{array}$
