Răspuns :

[tex]\sqrt{3x-1}-\sqrt{3x+1}>-1\\ \sqrt{3x-1}+1>\sqrt{3x+1}\\ (\sqrt{3x-1}+1)^2>(\sqrt{3x+1})^2\\ (\sqrt{3x-1})^2+2\cdot \sqrt{3x-1} \cdot 1+1^2>3x+1\\ 3x-1+2\cdot \sqrt{3x-1}+1>3x+1\\ 2\cdot \sqrt{3x-1}>1\\ \sqrt{3x-1}>\frac{1}{2}\\ (\sqrt{3x-1})^2>(\frac{1}{2})^2[/tex]
[tex]3x-1>\frac{1}{4}\\ 3x>\frac{1}{4}+1\\ 3x>\frac{5}{4}\\ x>\frac{5}{4}:3\\ x>\frac{5}{4} \cdot \frac{1}{3}[/tex]
[tex]x> \frac{5}{12}=>x \in ( \frac{5}{12},+ \infty) [/tex]