Đáp án:
$\begin{array}{l}
c)Theo\,a\\
\Rightarrow A\left( {\dfrac{{ - 2m}}{{m + 1}};0} \right);B\left( {0;2m} \right)\left( {m \ne - 1} \right)\\
\Rightarrow OA = \left| {\dfrac{{2m}}{{m + 1}}} \right|;OB = \left| {2m} \right|\\
\Rightarrow {S_{OAB}} = \dfrac{1}{2}.OA.OB\\
= \dfrac{1}{2}.\left| {\dfrac{{2m}}{{m + 1}}} \right|.\left| {2m} \right| = 2\\
\Rightarrow \dfrac{{4{m^2}}}{{2\left| {m + 1} \right|}} = 2\\
\Rightarrow {m^2} = \left| {m + 1} \right|\\
\Rightarrow \left[ \begin{array}{l}
{m^2} = m + 1\left( {khi:m > - 1} \right)\\
{m^2} = - m - 1\left( {khi:m < - 1} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{m^2} - m - 1 = 0\left( {m > - 1} \right)\\
{m^2} + m + 1 = 0\left( {vn} \right)
\end{array} \right.\\
\Rightarrow {m^2} - 2.m.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{5}{4} = 0\left( {m < - 1} \right)\\
\Rightarrow {\left( {m - \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\\
\Rightarrow m = \dfrac{{ - \sqrt 5 + 1}}{2}\left( {m < - 1} \right)\\
Vậy\,m = \dfrac{{1 - \sqrt 5 }}{2}
\end{array}$
