6.165. $\sqrt{\frac{18-7 x-x^{2}}{8-6 x+x^{2}}}+\sqrt{\frac{8-6 x+x^{2}}{18-7 x-x^{2}}}=\frac{13}{6}$.
## Solution.

Let's write the equation as

$$
\begin{aligned}
& \sqrt{-\frac{x^{2}+7 x-18}{x^{2}-6 x+8}}+\sqrt{-\frac{x^{2}-6 x+8}{x^{2}+7 x-18}}=\frac{13}{6} \Leftrightarrow \\
& \Leftrightarrow \sqrt{-\frac{(x+9)(x-2)}{(x-4)(x-2)}}+\sqrt{-\frac{(x-4)(x-2)}{(x+9)(x-2)}}=\frac{13}{6} \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array}{l}
\sqrt{-\frac{x+9}{x-4}}+\sqrt{-\frac{x-4}{x+9}}=\frac{13}{6} \\
x \neq 2
\end{array}\right.
\end{aligned}
$$

Domain of definition: $\left\{\begin{array}{l}-\frac{x+9}{x-4}>0, \\ x \neq 2,\end{array} \Leftrightarrow\left\{\begin{array}{l}(x+9)(x-4)<0, \\ x \neq 2,\end{array} \Leftrightarrow -9 < x < 4, x \neq 2\right.\right.$.

Let $y = \sqrt{-\frac{x+9}{x-4}}$. Then the equation becomes $y + \frac{1}{y} = \frac{13}{6} \Leftrightarrow 6 y^{2} - 13 y + 6 = 0$, from which $y_{1} = \frac{2}{3}, y_{2} = \frac{3}{2}$. Therefore, $\sqrt{-\frac{x+9}{x-4}} = \frac{2}{3}$ or $\sqrt{-\frac{x+9}{x-4}} = \frac{3}{2}$. This gives $-\frac{x+9}{x-4} = \frac{4}{9}$ or $-\frac{x+9}{x-4} = \frac{9}{4}$, from which $x_{1} = 0, x_{2} = -5$.

Answer: $x_{1} = 0, x_{2} = -5$.