Question

(4x-12)(x^2-9)=0

Answer 1

\(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=9\\4x=12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=3\end{matrix}\right.\)

Vậy \(S=\left\{\pm3\right\}\)

Answer 2

\(\left(4x-12\right)\left(x^2-9\right)=0\\ \begin{matrix}TH1:4x-12=0\\TH2:x^2-9=0\end{matrix}\)

Xét `TH1`

`4x-12=0`

`4x=12`

`x=3`

Xét `TH2:`

`x^2-9=0`

`x^2=9` ; `x^2=-9`

`x^2=3` ; `x^2=-3`

Vậy `TH1``x=3`

        `TH2x=`\(\pm3\)

`@An`