[tex](x+\sqrt{x^2-1})(x-\sqrt{x^2-1)} = 1
\\\;\\
\it x-\sqrt{x^2-1} =\dfrac{1}{x+\sqrt{x^2-1}}[/tex]
[tex]\it Notam\ \sqrt[2013]{x+\sqrt{x^2-1}} = t[/tex]
Ecuatia devine:
[tex]\it t+\dfrac{1}{t} =2 \Leftrightarrow t^2-2t+1=0\Leftrightarrow (t-1)^2=0 \Leftrightarrow t=1[/tex]
Deci:
[tex]\it \sqrt[2013]{x+\sqrt{x^2-1}} =1 \Longrightarrow x = 1[/tex]