(d1) y= ( 2m + 1 )x - ( 2m + 3)
(d2) y = (m-1)x + m
a, (d1) cắt (d2)⇔ 2m+1 $\neq$ m-1
⇔ 2m-m $\neq$ -1-1
⇔ m $\neq$ -2
Vậy m $\neq$ -2 thì (d1) cắt (d2
b, (d1) // (d2)
⇔$\left \{ {{2m+1=m-1} \atop {2m + 3 \neq m }} \right.$
⇔$\left \{ {{2m-m=-1-1} \atop {2m-m≠-3}} \right.$
⇔$\left \{ {{m=-2} \atop {m≠-3}} \right.$
Vậy m=-2 ; m≠ -3 thì (d1) // (d2)
c, (d1) ≡ (d2)
⇔$\left \{ {{2m+1=m-1} \atop {2m + 3 = m }} \right.$
⇔$\left \{ {{2m-m=-1-1} \atop {2m-m=-3}} \right.$
⇔$\left \{ {{m=-2} \atop {m=-3}} \right.$
Vậy m=-2 ; m= -3 thì (d1) ≡ (d2)
d, (d1) ⊥ (d2)
⇔(2m + 1).(m-1) =-1
⇔ 2m² -2m+m -1 =-1
⇔2m² -m =0
⇔ m (2m-1)=0
⇔ \(\left[ \begin{array}{l}m=0\\2m-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}m=0\\2m=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}m=0\\m=\frac{1}{2} \end{array} \right.\)
Vậy m =0 hoặc m = $\frac{1}{2}$ thì (d1) ⊥ (d2)
cuthilien
