6.052.

$$
\frac{1}{x-\sqrt{x^{2}-x}}-\frac{1}{x+\sqrt{x^{2}-x}}=\sqrt{3}
$$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x^{2}-x \geq 0, \\ x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x-1) \geq 0, \\ x \neq 0\end{array} \Leftrightarrow x \in(-\infty ; 0) \cup[1 ;+\infty)\right.\right.$.

From the condition we get

$$
\begin{aligned}
& \frac{x+\sqrt{x^{2}-x}-x+\sqrt{x^{2}-x}}{x^{2}-x^{2}+x}=\sqrt{3} \Leftrightarrow \frac{2 \sqrt{x^{2}-x}}{x}=\sqrt{3} \Leftrightarrow \\
& \Leftrightarrow \frac{2 \sqrt{x} \sqrt{x-1}}{x}=\sqrt{3} \Leftrightarrow \frac{2 \sqrt{x-1}}{\sqrt{x}}=\sqrt{3}(x \neq 0)
\end{aligned}
$$

Squaring both sides of the equation, we get $\frac{4 x-4}{x}=3$ or $x=4$. By verification, we confirm that $x=4$ is a root of the last equation with radicals.

Answer: $x=4$.