`3x^2-2x-1=0`
`<=>3x^2-3x+x-1=0`
`<=>3x(x-1)+(x-1)=0`
`<=>(x-1)(3x+1)=0`
`<=>` $\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\3x=-1\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{3}\end{array} \right.$
Vậy `S={1,-1/3}`
Ta có: \(3x^2-2x-1=0\)
\(\Leftrightarrow3x^2-3x+x-1=0\)
\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{-1}{3}\right\}\)
