Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)
=>\(t^2=x^2+3\Leftrightarrow x^2=t^2-3\)
Pt trở thành \(\left(3x+1\right)t=t^2-3+2x^2+2x+3\)
<=>\(t^2-\left(3x+1\right)+2x^2+2x=0\)
Có \(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=x^2-2x+1=\left(x-1\right)^2\)
Nên \(\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)
+, \(t=x+1\Leftrightarrow\sqrt{x^2+3}=x+1\Rightarrow x^2+3=x^2+2x+1\left(x\ge-1\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\left(TM\right)\)
+, \(t=2x\Leftrightarrow\sqrt{x^2+3}=2x\Rightarrow x^2+3=4x^2\left(x\ge0\right)\Leftrightarrow3x^2-3=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-1\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1;1\right\}\)
