6.029. $\frac{a^{2}-1}{a x-1}+\frac{a-x}{a}=1$.

6.029. $\frac{a^{2}-1}{a x-1}+\frac{a-x}{a}=1$.
## Solution.

Domain of definition: $a \neq 0, x \neq \frac{1}{a}$.

From the condition we have:

$$
\begin{aligned}
& \frac{a^{2}-1}{a x-1}+\frac{a}{a}-\frac{x}{a}=1 \Leftrightarrow \frac{a^{2}-1}{a x-1}+1-\frac{x}{a}=1 \Leftrightarrow \frac{a^{2}-1}{a x-1}-\frac{x}{a}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{a x^{2}-x-\left(a^{2}-1\right) a}{(a x-1) a}=0
\end{aligned}
$$

considering the domain of definition $a x^{2}-x-\left(a^{2}-1\right) a=0$, from which

$$
\begin{aligned}
& x_{1,2}=\frac{1 \pm \sqrt{1+4 a^{2}\left(a^{2}-1\right)}}{2 a}=\frac{1 \pm \sqrt{4 a^{4}-4 a^{2}+1}}{2 a}=\frac{1 \pm \sqrt{\left(2 a^{2}-1\right)^{2}}}{2 a}= \\
& =\frac{1 \pm\left(2 a^{2}-1\right)}{2 a} ; \\
& x_{1}=\frac{1-2 a^{2}+1}{2 a}=\frac{1-a^{2}}{a}, x_{2}=\frac{1+2 a^{2}-1}{2 a}=a
\end{aligned}
$$

Answer: $x_{1}=\frac{1-a^{2}}{a}, x_{2}=a$ for $a \neq 0$.