6.017. $\frac{x+n}{m+n}-\frac{m-n}{x-n}=\frac{x+p}{m+p}-\frac{m-p}{x-p}$.
## Solution.

Domain of definition: $x \neq n, x \neq p, m \neq -n, m \neq -p$.

From the condition, we get

$$
\begin{aligned}
& \frac{(x+n)(x-n)-(m-n)(m+n)}{(m+n)(x-n)}=\frac{(x+p)(x-p)-(m-p)(m+p)}{(m+p)(x-p)} \Leftrightarrow \\
& \Leftrightarrow \frac{x^{2}-n^{2}-m^{2}+n^{2}}{(m+n) x-n(m+n)}=\frac{x^{2}-p^{2}-m^{2}+p^{2}}{(m+p) x-p(m+p)} \Leftrightarrow \\
& \Leftrightarrow \frac{x^{2}-m^{2}}{(m+n) x-n(m+n)}=\frac{x^{2}-m^{2}}{(m+p) x-p(m+p)} \Leftrightarrow \\
& \Leftrightarrow \frac{x^{2}-m^{2}}{(m+n) x-n(m+n)}-\frac{x^{2}-m^{2}}{(m+p) x-p(m+p)}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}-m^{2}\right) \cdot\left(\frac{1}{(m+n) x-n(m+n)}-\frac{1}{(m+p) x-p(m+p)}\right)=0 .
\end{aligned}
$$

From this, we get $x^{2}-m^{2}=0 \Leftrightarrow x^{2}=m^{2}, x_{1,2}= \pm m$, or

$$
\begin{aligned}
& \frac{1}{(m+n) x-n(m+n)}-\frac{1}{(m+p) x-p(m+p)}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(m+p) x-p(m+p)-(m+n) x+n(m+n)}{((m+n) x-n(m+n))((m+p) x-p(m+p))}=0
\end{aligned}
$$

or, considering the domain of definition,

$$
\begin{aligned}
& (m+p) x-(m+n) x-p(m+p)+n(m+n)=0 \Leftrightarrow \\
& \Leftrightarrow(m+p) x-(m+n) x=p(m+p)-n(m+n) \Leftrightarrow \\
& \Leftrightarrow(m+p-m-n) x=p m+p^{2}-m n-n^{2} \Leftrightarrow(p-n) x=p^{2}-n^{2}+p m-m n \Leftrightarrow \\
& \Leftrightarrow(p-n) x=(p-n)(p+n)+m(p-n) \Leftrightarrow(p-n) x=(p-n)(p+n+m)
\end{aligned}
$$

## From this:

1) if $p-n=0, p=n$, then, considering the domain of definition, $x \in R$, except for $p$ and $n$;
2) if $p-n \neq 0, p \neq n$, then $x_{3}=p+n+m$.

Answer: if $n=p$, then $x \in R$, except for $n$ and $p$; if $n \neq p$, then $x_{1}=m$, $x_{2}=-m, x_{3}=m+n+p$.