Bài 1: Thiếu dữ kiện
Bài 2:
a, $( x+12).( x-3)= 0$
⇔ \(\left[ \begin{array}{l}x+12= 0\\x-3= 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x= -12\\x= 3\end{array} \right.\)
b, $( -x+5).( 3-x)= 0$
⇔ \(\left[ \begin{array}{l}-x+5= 0\\3-x= 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x= 5\\x= 3\end{array} \right.\)