6.028. $\frac{x}{a+b}+\frac{2a-x}{a-b}-\frac{a+b}{x}=1$.
Solution.

Domain of definition: $a \neq \pm b, x \neq 0$.

From the given, we have

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

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

Considering the domain of definition, $-2 b x^{2}+\left(a^{2}+2 a b+b^{2}\right) x-(a+b)^{2}(a-b)=0$ or $2 b x^{2}-(a+b)^{2} x+(a+b)^{2}(a-b)=0$, from which

$$
\begin{aligned}
& x_{1,2}=\frac{(a+b)^{2} \pm \sqrt{(a+b)^{4}-8 b(a+b)^{2}(a-b)}}{4 b}= \\
& =\frac{(a+b)^{2} \pm \sqrt{(a+b)^{2}\left((a+b)^{2}-8 b(a-b)\right)}}{4 b}= \\
& =\frac{(a+b)^{2} \pm(a+b) \sqrt{a^{2}+2 a b+b^{2}-8 a b+8 b^{2}}}{4 b}= \\
& =\frac{(a+b)^{2} \pm(a+b) \sqrt{a^{2}-6 a b+9 b^{2}}}{4 b}=\frac{(a+b)^{2} \pm(a+b) \sqrt{(a-3 b)^{2}}}{4 b}= \\
& =\frac{(a+b)^{2} \pm(a+b)(a-3 b)}{4 b}=\frac{(a+b)((a+b) \pm(a-3 b))}{4 b} \\
& x_{1}=\frac{(a+b)(a+b-a+3 b)}{4 b}=\frac{(a+b) 4 b}{4 b}=a+b, \\
& x_{2}=\frac{(a+b)(a+b+a-3 b)}{4 b}=\frac{(a+b)(2 a-2 b)}{4 b}=\frac{(a+b)(a-b)}{2 b}=\frac{a^{2}-b^{2}}{2 b}
\end{aligned}
$$

Answer: if $b \neq 0$, then $x_{1}=a+b, x_{2}=\frac{a^{2}-b^{2}}{2 b}$; if $b=0$, then $x=a$.