\(1+\sqrt{3x+1}=3x\)
<=> \(\sqrt{3x+1}=3x-1\)
<=> \(\begin{cases}3x-1\ge0\\3x+1=9x^2-6x+1\end{cases}\)
<=> \(\begin{cases}x\ge\frac{1}{3}\\x=\frac{3\pm\sqrt{5}}{6}\end{cases}\)
=> x=\(\frac{3+\sqrt{5}}{6}\)
\(1+\sqrt{3x+1}=3x\)
<=> \(\sqrt{3x+1}=3x-1\)
<=> \(\left[\begin{array}{nghiempt}3x-1\ge0\\3x+1=9x^2-6x+1\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x\ge\frac{1}{3}\\9x-9x^2=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x\ge\frac{1}{3}\\9x\left(1-x\right)=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x\ge\frac{1}{3}\\\left[\begin{array}{nghiempt}x=0\left(L\right)\\x=1\left(N\right)\end{array}\right.\end{array}\right.\) => \(x=1\)
S=\(\left\{1\right\}\)
