\(\begin{array}{l}
a)\,Thay\,x = 0;y = 0\,vao\,ham\,so\,ta\,duoc:\\
0 = m.0 - 2m - 1 \Rightarrow m = - \frac{1}{2}\\
b)\,DK:\,m \ne 0\\
d\,cat\,Ox\,tai\,A\left( {\frac{{2m + 1}}{m};0} \right);B\left( {0; - 2m - 1} \right)\\
+ )\,OAB\,can \Rightarrow \left| {\frac{{2m + 1}}{m}} \right| = \left| { - 2m - 1} \right|\\
\Leftrightarrow \left| {2m + 1} \right| = \left| m \right|\left| {2m + 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {2m + 1} \right| = 0\\
\left| m \right| = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = - \frac{1}{2}\\
m = 1\\
m = - 1
\end{array} \right.\\
c)\,{S_{OAB}} = \frac{1}{2}\left| {\frac{{2m + 1}}{m}} \right|.\left| { - 2m - 1} \right|\\
= \frac{1}{2}\frac{{{{\left( {2m + 1} \right)}^2}}}{{\left| m \right|}} = 4,5\\
\Leftrightarrow \frac{{{{\left( {2m + 1} \right)}^2}}}{{\left| m \right|}} = 9\\
\Rightarrow {\left( {2m + 1} \right)^2} = 9\left| m \right|\\
\Leftrightarrow \left[ \begin{array}{l}
4{m^2} + 4m + 1 = 9m\\
4{m^2} + 4m + 1 = - 9m
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = \dfrac{1}{5}\\
m = \dfrac{{ - 13 \pm 3\sqrt {17} }}{8}
\end{array} \right.
\end{array}\)
