\( (d)//(d')\\→\begin{cases}m=\sqrt{m+5}-1\\3\ne 3(vô\,\,lý)\end{cases}\)
\(→(d)≡(d')\\m=\sqrt{m+5}-1\\↔\sqrt{m+5}=m+1\\↔m+5=m^2+2m+1\\↔-m^2-m+4=0\\↔m^2+m-4=0\\Δ=1^2-4.1.(-4)=15>0\)
\(→\) Pt có 2 nghiệm phân biệt
\(m_1=\dfrac{-1+\sqrt{17}}{2}(TM)\\m_2=\dfrac{-1-\sqrt{17}}{2}(KTM)\)
Vậy \(m=\dfrac{-1+\sqrt{17}}{2}\)