Question

Radical (3x - 1) - radical (3x + 1) > - 1

Answer 1

$\sqrt{3x-1} - \sqrt{3x+1}>-1\\ \\ \sqrt{3x+1} - \sqrt{3x-1} < 1 \\ \\f:\Big[\dfrac{1}{3},+\infty\Big)\to \mathbb{R},\quad f(x) = \sqrt{3x+1}-\sqrt{3x-1}\\ \\ \text{Functia e strict descrescatoare pe }\Big[\dfrac{1}{3},+\infty\Big)= D_{max}, \\ \text{deoarece }f(a) > f(b),\quad \forall a < b. \\\\ \text{Astfel}\,\,\,f(x) < 1 \Rightarrow x > f^{-1}(1)$

$x = \sqrt{3y-1}-\sqrt{3y+1} \\ \\x = 1 \Rightarrow \sqrt{3y+1}-\sqrt{3y-1} = 1 \\\\ \Rightarrow 3y+1 +(3y-1) -2\sqrt{9y^2-1} = 1\\ \Rightarrow 6y-2\sqrt{9y^2-1} = 1 \\ \Rightarrow 2\sqrt{9y^2-1} = 6y-1\\ \Rightarrow 4(9y^2-1) = 36y^2 - 12y+1 \\ \Rightarrow 12y =5 \Rightarrow y = \dfrac{5}{12} \Rightarrow f^{-1}(1) = \dfrac{5}{12}\\ \\ \\ \Rightarrow x > \dfrac{5}{12} \Rightarrow \boxed{x\in \Big(\dfrac{5}{12},+\infty\Big)}$