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

Domain of definition: $x \neq 1$.

We bring the equation to a common denominator:

$$
\frac{a x^{2}-(a+1)^{2}(x-1)}{x-1}=0 \Leftrightarrow \frac{a x^{2}-(a+1)^{2} x+(a+1)^{2}}{x-1}=0
$$

Considering the domain of definition, $a x^{2}-(a+1)^{2} x+(a+1)^{2}=0$, from which

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

Answer: $x_{1}=\frac{a+1}{a}$, where $a \neq 0 ; x_{2}=a+1$.