262using (x+\sqrt{x^{2}+1} ) = \frac{1}{\sqrt{x^{2}+1} -x}
we get x=±y
\hspace{-16}$If $\mathbf{x,y\in\mathbb{R^{*}}}$ and $\mathbf{xy\leq 1}$ and $\mathbf{\left(x+\sqrt{x^2+1}\right).\left(y+\sqrt{y^2+1}\right)=1}$\\\\\\ Then $\mathbf{\sqrt{\frac{x}{y}+\frac{y}{x}+6}=}$
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