\(\frac{2x}{\sqrt{2x+1}-1}>2x+2\) ( ĐK : \(x\ge-\frac{1}{2};x\ne0\) )
\(\Leftrightarrow\frac{2x}{\frac{2x}{\sqrt{2x+1}+1}}>2x+2\)
\(\Leftrightarrow\sqrt{2x+1}+1>2x+2\)
\(\Leftrightarrow\sqrt{2x+1}>2x+1\)
Với \(x=-\frac{1}{2}\) . BPT ko thỏa mãn
Với \(x>-\frac{1}{2}\) .
\(\Leftrightarrow2x+1>4x^2+4x+1\)
\(\Leftrightarrow-4x^2-2x>0\)
\(\Leftrightarrow-\frac{1}{2}< x< 0\)
Vậy \(S=\left(-\frac{1}{2};0\right)\)