\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\left(x+2\right)^2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\dfrac{4\left(x+2\right)^2}{4}\)
\(\Leftrightarrow\left(x-2\right)^3=4\left(x+2\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=4\left(x^2+4x+4\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=4x^2+16x+16\)
\(\Leftrightarrow x^3-10x^2-4x-24=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\left(x+2\right)^2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\dfrac{4\left(x+2\right)^2}{4}\)
\(\Leftrightarrow\left(x-2\right)^3=4\left(x+2\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=4\left(x^2+4x+4\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=4x^2+8x+8\)
\(\Leftrightarrow x^3-10x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-10x+4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-10x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=5+\sqrt{21}\\x=5-\sqrt{21}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{0;5+\sqrt{21};5-\sqrt{21}\right\}\)
