a ) (x+1)(3x-1)=0
<=> \(\left[ \begin{array}{l}x+1=0\\3x-1=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-1\\x=\frac{1}{3} \end{array} \right.\)
Vậy S = {-1; $\frac{1}{3}$ }
b ) (3x-16) (2x-3) = 1
<=> (3x-16)(2x-3) -1 = 0
<=> (3x-16)(-2x+3) = 0
<=> \(\left[ \begin{array}{l}3x-16 = 0\\-2x+3=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\frac{16}{3} \\x=\frac{3}{2} \end{array} \right.\)
Vậy S = { $\frac{16}{3}$ ;$\frac{3}{2}$ }
#number one
#Trang Huyen
