a.(4x-2).(x+5)=0
⇔\(\left[ \begin{array}{l}4x-2=0\\x+5=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}4x=2\\x=-5\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-5\end{array} \right.\)
Vậy x=$\frac{1}{2}$; x=-5
b.72.(28-49)+28.(-49-72)
=72.28-72.49-28.49-28.72
=(72.28-28.49-28.72)-72.49
=28(72-49-72)-72.49
=-28.49-72.49
=-49(72+28)
=-49.100
=-4900.