Đáp án:
$\begin{array}{l}
a)\\
+ Cho:x = 0 \Rightarrow y = 2m \Rightarrow B\left( {0;2m} \right)\\
+ Cho:y = 0 \Rightarrow x = \dfrac{{ - 2m}}{{m + 1}}\left( {m \ne - 1} \right)\\
\Rightarrow A\left( {\dfrac{{ - 2m}}{{m + 1}};0} \right)\\
\Rightarrow \left\{ \begin{array}{l}
OA = \left| {\dfrac{{ - 2m}}{{m + 1}}} \right| = \left| {\dfrac{{2m}}{{m + 1}}} \right|\\
OB = \left| {2m} \right|
\end{array} \right.\\
OA = OB\\
\Rightarrow \left| {\dfrac{{2m}}{{m + 1}}} \right| = \left| {2m} \right|\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{2m}}{{m + 1}} = 2m\\
\dfrac{{2m}}{{m + 1}} = - 2m
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2m.\left( {\dfrac{1}{{m + 1}} - 1} \right) = 0\\
2m.\left( {\dfrac{1}{{m + 1}} + 1} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = 0\\
m + 1 = 1\\
m + 1 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = 0\left( {tm} \right)\\
m = - 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 0;m = - 2
\end{array}$
