\(pt\Leftrightarrow x^3-3x^2-4x+12=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-2;2;3\right\}\)
x3+12=3x2+4x
⇔x3-3x2-4x+12=0
⇔(x3-3x2)-(4x-12)=0
⇔x2(x-3)-4(x-3)=0
⇔(x-3)(x2-4)=0
⇔(x-3)(x-2)(x+2)=0
=>\(\left\{{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)
vậy S{-2,2,3}
