$$n\ge2\implies x_n=\frac{1}{\sqrt[2^{n-1}]{2}-1}$$

Solution: 1) Squaring, we get: $x_{n}^{2}-2{{x}_{n}}{{x}_{n-1}}-{{x}_{n-1}}=0$ 2) Now let ${{x}_{n}}=\frac{1}{{{y}_{n}}}$, and we get $y_{n}^{2}+2{{y}_{n}}={{y}_{n-1}}$ 3) Now let: ${{y}_{n}}+1={{z}_{n}}\Rightarrow z_{n}^{2}={{z}_{n-1}}$. 4) Step by step with induction, we get: ${{z}_{n}}=\sqrt[{{2}^{n-1}}]{2}\Rightarrow {{x}_{n}}=\frac{1}{\sqrt[{{2}^{n-1}}]{2}-1},\forall n\ge 2$

I'm a fool...