\(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow3x+1=0\) hay \(-x-2=0\)
\(\Leftrightarrow x=\dfrac{-1}{3}\) hay \(x=-2\)
-Vậy \(S=\left\{\dfrac{-1}{3};-2\right\}\)
\(\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\Leftrightarrow\left(3x+1\right)\left(-x-2\right)=0\Leftrightarrow x=-\dfrac{1}{3};x=-2\)
=>(3x)2-1=(3x+1)(4x+1)
=>(3x-1)(3x+1)=(3x+1)(4x+1)
=>(3x-1)(3x+1)-(3x+1)(4x+1)=0
=>(3x+1).(3x-1-4x-1)=0
=>3x+1=0 hoặc -x-2=0
=>x=-1/3 x=-2
\(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};-2\right\}\)