\(x=0→y=1→OB=|1|=1\\y=0→x=-\dfrac{1}{m}→OA=\bigg|-\dfrac{1}{m}\bigg|\)
Để \(ΔOAB\) vuông cân tại \(O\)
\(→OA=B\) hay \(\bigg|-\dfrac{1}{m}\bigg|=1(m\ne 0)\\↔\left[\begin{array}{1}-\dfrac{1}{m}=1\\-\dfrac{1}{m}=-1\end{array}\right.\\↔\left[\begin{array}{1}m=-1\\m=1\end{array}\right.\)
Vậy \(m∈\{-1;1\}\)