\(x^2-1=3\sqrt{3x+1}\)
\(ĐK:x\ge-\dfrac{1}{3}\);\(x^2-1\ge0\)
\(\Leftrightarrow x^2-1-3\sqrt{3x+1}=0\)
\(\Leftrightarrow x^2-1+3x+1+\dfrac{9}{4}-\left(3x+1+3\sqrt{3x+1}+\dfrac{9}{4}\right)=0\)
\(\Leftrightarrow\left(x^2+3x+\dfrac{9}{4}\right)-\left(3x+1+3\sqrt{3x+1}+\dfrac{9}{4}\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2-\left(\sqrt{3x+1}+\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\left(\sqrt{3x+1}+\dfrac{3}{2}\right)^2\)
`@`TH1:\(x+\dfrac{3}{2}=\sqrt{3x+1}+\dfrac{3}{2}\)
\(\Leftrightarrow x^2=3x+1\)
\(\Leftrightarrow x^2-3x+1=0\)
\(\Delta=\left(-3\right)^2-4=9-4=5>0\)
\(\rightarrow\left\{{}\begin{matrix}x=\dfrac{3+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{3-\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)
`@`TH2:\(-x-\dfrac{3}{2}=\sqrt{3x+1}+\dfrac{3}{2}\)
\(\Leftrightarrow\sqrt{3x+1}+x+3=0\) ( vô lý )
Vậy \(S=\left\{\dfrac{3+\sqrt{5}}{2}\right\}\)
