\( (d)\) cắt \(Ox\) tại \(A\)
\(→(d)\) cắt \( Ox\) tại \(A(x;0)\)
\(→(m-1)x+4=0\\↔x=\dfrac{4}{1-m}\\→OA=\bigg|\dfrac{4}{1-m}\bigg|(m\ne 1)\)
\( (d)\) cắt \( Oy\) tại \(B\)
\(→(d)\) cắt \(Oy\) tại \(B(0;y)\)
\(→(m-1).0+4=y\\↔y=4\\→OB=|4|=4\)
\(S_{ΔOAB}=2\\→\dfrac{1}{2}.OA.OB=2\\↔\dfrac{1}{2}.\bigg|\dfrac{4}{1-m}\bigg|.4=2\\↔\bigg|\dfrac{4}{1-m}\bigg|=1\\↔\left[\begin{array}{1}\dfrac{4}{1-m}=4\\\dfrac{4}{1-m}=-1\end{array}\right.\\↔\left[\begin{array}{1}1-m=4\\1-m=-4\end{array}\right.\\↔\left[\begin{array}{1}m=-3(TM)\\m=5(TM)\end{array}\right.\)
Vậy \(m∈\{-3;5\}\)
