9x2 - 1 = (3x + 1)(4x + 1)
<=> (3x - 1)(3x + 1) - (3x + 1)(4x + 1) = 0
<=> (3x + 1)(3x - 1 - 4x - 1) = 0
<=> (3x + 1)(- 2 - x) = 0
<=> \(\left[\begin{matrix}3x+1=0\\-2-x=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy S = { - 1/3 ; -2}
\(9x^2-1=(3x+1)(4x+1)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(4x+1\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
