[tex]\sqrt{x+1}-\sqrt{x-1} = \sqrt2 \\ \\ $Notam:\quad $ a = \sqrt{x+1};\quad b = \sqrt{x-1} \\ $Facem sistem: $ \\ \\\left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = x+1-(x-1) \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = x+1-x+1 \end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a^2-b^2 = 2 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ (a-b)(a+b) = 2 \end{array} \right \Rightarrow [/tex]
[tex]\Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ \sqrt2\cdot (a+b) = 2 \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{2}{\sqrt2} \end{array} \right \Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{\sqrt2\cdot 2}{\sqrt2\cdot \sqrt2} \end{array} \right \Rightarrow \\ \\ [/tex]
[tex]\Rightarrow \left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b = \dfrac{\sqrt2\cdot 2}{2} \end{array} \right \Rightarrow\left\{ \begin{array}{c} a-b = \sqrt2 \\ a+b =\sqrt2}\end{array} \right |_{(adunam)} \Rightarrow \\ \\ \Rightarrow a+a-b+b = \sqrt2+\sqrt2 \Rightarrow 2a = 2\sqrt2 \Rightarrow a = \sqrt2 \Rightarrow \\ \\ \Rightarrow \sqrt{x+1} = \sqrt2 \Big|^2 \Rightarrow x+1 = 2 \Rightarrow \boxed{x = 1} \\ \\ [/tex]
[tex]a-b = \sqrt2 \Rightarrow \sqrt2-b = \sqrt2 \Rightarrow b = 0 \Rightarrow \sqrt{x-1} = 0 \Rightarrow x-1 = 0\Rightarrow \\ \\ \Rightarrow \boxed{x=1} \\ \\ \\ \Rightarrow \boxed{S = \Big\{1\Big\}}[/tex]
