y(y2-1) = y2 - 5y + 6 = 0
⇔$\left \{ {{y(y-1)(y+1)=0} \atop {y^2-2y-3y+6=0}} \right.$
⇔$\left \{ {{y(y-1)(y+1)=0} \atop {y(y-2)-3(y-2)=0}} \right.$
⇔$\left \{ {{y(y-1)(y+1)=0} \atop {(y-2)(y-3)=0}} \right.$
⇔$\left \{ {{y=0;y=1;y=-1} \atop {y=2;y=3}} \right.$
⇒ PT vô nghiệm
