6.160. $\sqrt{\frac{20+x}{x}}+\sqrt{\frac{20-x}{x}}=\sqrt{6}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\frac{20+x}{x} \geq 0, \\ \frac{20-x}{x} \geq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x+20) \geq 0, \\ x(x-20) \geq 0, \\ x \neq 0,0<x \leq 20 .\end{array}\right.\right.$

By squaring both sides of the equation, we have

$$
\begin{aligned}
& \frac{20+x}{x}+2 \sqrt{\frac{(20+x)(20-x)}{x^{2}}}+\frac{20-x}{x}=6 \Leftrightarrow \frac{\sqrt{400-x^{2}}}{x}=\frac{3 x-20}{x} \Rightarrow \\
& \sqrt{400-x^{2}}=3 x-20 \text {, where } 3 x-20 \geq 0, x \geq \frac{20}{3} .
\end{aligned}
$$

Further: $400-x^{2}=9 x^{2}-120 x+400 \Leftrightarrow x(x-12)=0$, from which $x_{1}=0$, $x_{2}=12 ; x_{1}=0<\frac{20}{3}$ does not fit.

Answer: $x=12$.