\(m=4→x^2-4x+4-1\\↔x^2-3x-x+3=0\\↔x(x-3)-(x-3)=0\\↔(x-1)(x-3)=0\\↔\left[\begin{array}{1}x-1=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=1\\x=3\end{array}\right.\\Δ=(-m)^2-4.1.(m-1)=m^2-4m+4=(m-4)^2\\\text{Pt có nghiệm kép}\,\,→Δ=(m-4)^2=0↔m-4=0↔m=4\\\text{Pt có 2 nghiệm phân biệt}\,\,→Δ=(m-4)^2>0\\↔\left[\begin{array}{1}m-4>0\\m-4<0\\m-4\ne 0\end{array}\right.\\↔\left[\begin{array}{1}m>4\\m<4\\m\ne 0\end{array}\right.\\\text{Pt vô nghiệm}\,\,→Δ=(m-4)^2<0(vô\,\,lý)→m∈varnothing\)