\(y'=-3x^2+3m=0\Rightarrow x^2=m\)
Hàm có 2 cực trị khi \(m>0\Rightarrow\left[{}\begin{matrix}x=\sqrt{m}\\x=-\sqrt{m}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(\sqrt{m};2m\sqrt{m}+1\right)\\B\left(-\sqrt{m};-2m\sqrt{m}+1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{OA}=\left(\sqrt{m};2m\sqrt{m}+1\right)\\\overrightarrow{OB}=\left(-\sqrt{m};-2m\sqrt{m}+1\right)\end{matrix}\right.\)
\(OA\perp OB\Rightarrow\overrightarrow{OA}.\overrightarrow{OB}=0\)
\(\Rightarrow-m+\left(1-4m^3\right)=0\)
\(\Leftrightarrow\left(2m-1\right)\left(2m^2+m+1\right)=0\)
\(\Leftrightarrow m=\dfrac{1}{2}\)
