Đáp án:
$\begin{array}{l}
- Cho:x = 0 \Rightarrow y = 3m + 4\\
\Rightarrow B\left( {0;3m + 4} \right)\\
\Rightarrow OB = \left| {3m + 4} \right|\\
- Cho:y = 0 \Rightarrow x = \dfrac{{ - 3m - 4}}{{4m - 3}}\left( {m \ne \dfrac{3}{4}} \right)\\
\Rightarrow A\left( {\dfrac{{ - 3m - 4}}{{4m - 3}};0} \right)\\
\Rightarrow OA = \left| {\dfrac{{ - 3m - 4}}{{4m - 3}}} \right| = \left| {\dfrac{{3m + 4}}{{4m - 3}}} \right|\\
\Delta OAB\,can\\
\Rightarrow OA = OB\\
\Rightarrow \left| {3m + 4} \right| = \left| {\dfrac{{3m + 4}}{{4m - 3}}} \right|\\
\Rightarrow \left[ \begin{array}{l}
3m + 4 = \dfrac{{3m + 4}}{{4m - 3}}\\
3m + 4 = - \dfrac{{3m + 4}}{{4m - 3}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left( {3m + 4} \right).\left( {\dfrac{1}{{4m - 3}} - 1} \right) = 0\\
\left( {3m + 4} \right).\left( {\dfrac{1}{{4m - 3}} + 1} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = - \dfrac{4}{3}\\
4m - 3 = 1\\
4m - 3 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = - \dfrac{4}{3}\\
m = 1\\
m = \dfrac{1}{2}
\end{array} \right.\left( {tm} \right)\\
Vậy\,m = - \dfrac{4}{3};m = \dfrac{1}{2};m = 1
\end{array}$
