Ta có: |x|-|2x+3|=x-1
\(\Leftrightarrow\left(\left|x\right|-\left|2x+3\right|\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\left(\left|x\right|\right)^2-2\cdot\left|x\right|\cdot\left|2x+3\right|+\left(\left|2x+3\right|\right)^2=x^2-2x+1\)
\(\Leftrightarrow\left(-x\right)^2-2\cdot\left(-x\right)\cdot\left(2x+3\right)+\left(2x+3\right)^2=x^2-2x+1\)
\(\Leftrightarrow x^2+2x\left(2x+3\right)+\left(2x+3\right)^2=x^2-2x+1\)
\(\Leftrightarrow x^2+4x^2+6x+4x^2+12x+9=x^2-2x+1\)
\(\Leftrightarrow9x^2+18x+9-x^2+2x-1=0\)
\(\Leftrightarrow8x^2+20x+8=0\)
\(\Leftrightarrow8x^2+16x+4x+8=0\)
\(\Leftrightarrow8x\left(x+2\right)+4\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\8x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\8x=-4\end{matrix}\right.\Leftrightarrow x=-\frac{1}{2}\)
Vậy: \(S=\left\{-\frac{1}{2}\right\}\)
