Đáp án:

\(\begin{array}{l}
a)\,\,m =  - \frac{1}{2}\\
b)\,\,\,m =  \pm 1\,\,\,\,thi\,\,\,\Delta AOB\,\,\,can.\\
m \in \left\{ {\frac{{ - 13 - 3\sqrt {17} }}{8};\,\,\frac{{ - 13 + 3\sqrt {17} }}{8};\,\,\frac{1}{4};\,\,1} \right\}\,\,\,thi\,\,\,{S_{OAB}} = \frac{9}{2}.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
d:\,\,\,y = mx - 2m - 1\,\,\,\,\left( {m \ne 0} \right)\\
a)\,\,\,\,Do\,\,\,thi\,\,\,hs\,\,\,di\,\,\,qua\,\,\,goc\,\,\,toa\,\,\,do\\
 \Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
 - 2m - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
m =  - \frac{1}{2}
\end{array} \right. \Leftrightarrow m =  - \frac{1}{2}.\\
b)\,\,\,d\,\,\,\,cat\,\,\,cac\,\,\,truc\,\,\,toa\,\,\,do \Rightarrow  - 2m - 1 \ne 0 \Leftrightarrow m \ne  - \frac{1}{2}\,\,\\
Ta\,\,co:\,\,\,d \cap Ox = \left\{ A \right\} \Rightarrow A\left( {\frac{{2m + 1}}{m};\,\,\,0} \right)\\
d \cap Oy = \left\{ B \right\} \Rightarrow B\left( {0; - 2m - 1} \right).\\
 + )\,\,\,\Delta AOB\,\,\,la\,\,\,tam\,\,giac\,\,can \Rightarrow \Delta AOB\,\,\,can\,\,\,tai\,\,O\\
 \Rightarrow OA = OB \Leftrightarrow \left| {{x_A}} \right| = \left| {{y_B}} \right|\\
 \Leftrightarrow \left| {\frac{{2m + 1}}{m}} \right| = \left| { - 2m - 1} \right| \Leftrightarrow \frac{{\left| {2m + 1} \right|}}{{\left| m \right|}} = \left| {2m + 1} \right|\\
 \Leftrightarrow \left| {2m + 1} \right|\left( {\frac{1}{{\left| m \right|}} - 1} \right) = 0\\
 \Leftrightarrow 1 - \left| m \right| = 0\,\,\,\,\,\left( {do\,\,\,2m + 1 \ne 0} \right)\\
 \Leftrightarrow \left| m \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
m = 1\,\,\,\left( {tm} \right)\\
m =  - 1\,\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,\,m =  \pm 1\,\,\,\,thi\,\,\,\Delta AOB\,\,\,can.\\
 + )\,\,\,\Delta AOB\,\,co\,\,\,S = 4,5\\
 \Leftrightarrow \frac{1}{2}OA.OB = \frac{9}{2}\\
 \Leftrightarrow OA.OB = 9\\
 \Leftrightarrow \left| {{x_A}} \right|.\left| {{y_B}} \right| = 9\\
 \Leftrightarrow \left| {\frac{{2m + 1}}{m}} \right|.\left| { - 2m - 1} \right| = 9 \Leftrightarrow \frac{{\left| {2m + 1} \right|}}{{\left| m \right|}}.\left| {2m + 1} \right| = 9\\
 \Leftrightarrow {\left( {2m + 1} \right)^2} = 9\left| m \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
{\left( {2m + 1} \right)^2} = 9m
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
{\left( {2m + 1} \right)^2} =  - 9m
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
4{m^2} - 5m + 1 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
4{m^2} + 13m + 1 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m = 1\\
m = \frac{1}{4}
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
\left[ \begin{array}{l}
m = \frac{{ - 13 + 3\sqrt {17} }}{8}\\
m = \frac{{ - 13 - 3\sqrt {17} }}{8}
\end{array} \right.
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = \frac{1}{4}\\
m = \frac{{ - 13 + 3\sqrt {17} }}{8}\\
m = \frac{{ - 13 - 3\sqrt {17} }}{8}
\end{array} \right.\\
Vay\,\,m \in \left\{ {\frac{{ - 13 - 3\sqrt {17} }}{8};\,\,\frac{{ - 13 + 3\sqrt {17} }}{8};\,\,\frac{1}{4};\,\,1} \right\}\,\,\,thi\,\,\,{S_{OAB}} = \frac{9}{2}.
\end{array}\)