ĐK: \(x\ge0\),x\(\le-1\)
\(3x^2+x-2\sqrt{x^2+x}=1\Leftrightarrow3\left(x^2+x\right)-2\sqrt{x^2+x}=1\Leftrightarrow3\left(x^2+x\right)-2\sqrt{x^2+x}-1=0\)(*)
Đặt a=\(\sqrt{x^2+x}\left(a\ge0\right)\)
Vậy (*)\(\Leftrightarrow3a^2-2a-1=0\Leftrightarrow\left(a-1\right)\left(3a+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-1=0\\3a+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=-\dfrac{1}{3}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a=1\Leftrightarrow\sqrt{x^2+x}=1\Leftrightarrow x^2+x=1\Leftrightarrow x^2+x-1=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{-1-\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy S={\(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\)}
