[tex]\sqrt[2017]{x+\sqrt{x^2-1}}+\sqrt[2017]{x-\sqrt{x^2-1}} = 2\\ \\ a = \sqrt[2017]{x+\sqrt{x^2-1}}\\ b = \sqrt[2017]{x-\sqrt{x^2-1}} \\ \\ ab = \sqrt[2017]{(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})}} =\sqrt[2017]{x^2-(x^2-1)} = 1\\ \\ a+b = 2\Big|^2 \Rightarrow a^2+2ab+b^2= 4 \Rightarrow a^2+2+b^2 = 4 \Rightarrow \\ \\\Rightarrow a^2+b^2 = 2 \\ \\ \begin{cases} a+b = 2 \\ a^2+b^2 = 2\end{cases}\Bigg| \Rightarrow (a = 0\,\text{ si }\,b = 0)\quad \text{sau}\quad (a=1\,\text{ si }\,{ b = 1})[/tex]
[tex](a = 0\,\text{ si }\,b = 0)\quad (F) \\ \\ \text{Deoarece }x\pm\sqrt{x^2-1} = 0 \Rightarrow x = \mp\sqrt{x^2-1} \Rightarrow x^2 = x^2-1\, (F) \\ \\ (a=1\, \text{ si }\, b=1) \\ \\ \Rightarrow \sqrt[2017]{x-\sqrt{x^2-1}} = 1\Big|^{2017} \Rightarrow x-\sqrt{x^2-1} = 1 \Rightarrow \\ \\ \Rightarrow \sqrt{x^2-1} = x-1 \Rightarrow x^2-1 = x^2-2x+1 \Rightarrow 2x = 2 \Rightarrow \boxed{x = 1} \\ \\ \Rightarrow \sqrt[2017]{x-\sqrt{x^2-1}} = 1 \Rightarrow \text{ (... ...) }\Rightarrow \boxed{x = 1}[/tex]
[tex]\Rightarrow M = \{1\} \cap \{1\} \Rightarrow M = \{1\} \\ \\ \Rightarrow \displaystyle \sum\limits_{x\in M} x^2 = 1^2 = \boxed{1}[/tex]
