$11.14 \sqrt{x^{2}-1}-\frac{6}{\sqrt{x^{2}-1}}=1$.
11.14 Here $\quad x^{2}-1>0 \Rightarrow|x|>1 \Rightarrow x1$. Let $\sqrt{x^{2}-1}=y>0$. Further, we have $y-\frac{6}{y}=1 ; y^{2}-y-6=0$; $y_{1}=-2<0$ (not suitable), $y_{2}=3$. Thus, we obtain the equation $\sqrt{x^{2}-1}=3$, from which $x^{2}-1=9$, i.e., $x_{1,2}= \pm \sqrt{10}$. Both roots belong to the domain of the equation. Answer: $x_{1,2}= \pm \sqrt{10}$.