`\text{~~Holi~~}`
`(8x-4)(x^2+2x+2)=0`
`->`\(\left[ \begin{array}{l}8x-4=0\\x^2+2x+2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}8x=4\\x^2+2x+1+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\(x+1)^2+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\(x+1)^2=-1 (\text{vô nghiệm, vì} (x+1)^2\ge0∀x)\end{array} \right.\)
`-> x=1/2`
Vậy `S={1/2}`
