$(d)//(d')$
$→\begin{cases}2m^2+3m+1=6 \\m-2\ne -1\end{cases}$
$(*)2m^2+3m+1=6\\↔2m^2+3m-5=0\\↔2m^2+5m-2m-5=0\\↔(2m^2+5m)-(2m+5)=0\\↔m(2m+5)-(2m+5)=0\\↔(m-1)(2m+5)=0\\↔\left[\begin{array}{1}m-1=0\\2m+5=0\end{array}\right.\\↔\left[\begin{array}{1}m=1\\2m=-5\end{array}\right.\\↔\left[\begin{array}{1}m=1\\m=-\dfrac{5}{2}\end{array}\right.$
$(*)m-2\ne -1\\↔m\ne 1$
Kết hợp 2 dấu $(*)$
$→m=-\dfrac{5}{2}$
Vậy $m=-\dfrac{5}{2}$ thì $(d)//(d')$
