ĐK: 2x+1≥0⇔2x≥-1⇒x≥$\frac{-1}{2}$
Ta có: |-4x|=2x+1
⇔ \(\left[ \begin{array}{l}-4x=2x+1\\-4x=-2x-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}-6x=1\\-2x=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-1}{6}(TMĐK) \\x=\frac{1}{2}(TMĐK)\end{array} \right.\)
Vậy s={\frac{-1}{6}; \frac{1}{2}}