a/ ( x + 1 )( 3 - x ) = 0
⇔ \(\left[ \begin{array}{l}x+1=0\\3-x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1.\\x=3.\end{array} \right.\)
Vậy...
---
b/ ( x - 2 )( 2x - 1 ) = 0
⇔ \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2.\\2x=1.( vô lí )\end{array} \right.\)
Vậy...
---
c/ ( 3x + 9 )( 1 - 3x ) = 0
⇔ \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=-9\\3x=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-3.\\x=1:3(vô lí )\end{array} \right.\)
Vậy...
