`3x^2-5x-12=0`
`<=>3x^2-9x+4x-12=0`
`<=>3x(x-3)+4(x-3)=0`
`<=>(x-3)(3x+4)=0`
`<=>[(x-3=0),(3x+4=0):}<=>[(x=3),(x=-4/3):}`
`3x^2 -5x-12=0`
`<=>3x^2 -9x+4x-12=0`
`<=>3x(x-3)+4(x-3)=0`
`<=>(x-3)(3x+4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\\ \Leftrightarrow\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy `S={3; -4/3}`
\(3x^2-5x-12=0\)
\(\Leftrightarrow\left(3x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{4}{3};3\right\}\).
