Đáp án:
$\begin{array}{l}
+ Cho:x = 0 \Rightarrow y = \dfrac{{ - m}}{9}\\
\Rightarrow A\left( {0;\dfrac{{ - m}}{9}} \right) \in Oy\\
\Rightarrow OA = \left| {\dfrac{{ - m}}{9}} \right| = \dfrac{m}{9}\\
+ Cho:y = 0\\
\Rightarrow {x^2} = \dfrac{m}{9}\left( {dk:m > 0} \right)\\
\Rightarrow x = \dfrac{{\sqrt m }}{3}/x = \dfrac{{ - \sqrt m }}{3}\\
\Rightarrow B\left( {\dfrac{{\sqrt m }}{3};0} \right);C\left( { - \dfrac{{\sqrt m }}{3};0} \right)\\
\Rightarrow BC = \dfrac{{2\sqrt m }}{3}\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.OA.BC\\
= \dfrac{1}{2}.\dfrac{m}{9}.\dfrac{{2\sqrt m }}{3}\\
= \dfrac{{m\sqrt m }}{{27}} = 8\\
\Rightarrow m\sqrt m = 27.8\\
\Rightarrow {\left( {\sqrt m } \right)^3} = {\left( 6 \right)^3}\\
\Rightarrow \sqrt m = 6\\
\Rightarrow m = 36\left( {tmdk} \right)
\end{array}$
Vậy m=36
