Three. (Full marks 12 points) Solve the equation:
$$
\frac{13 x-x^{2}}{x+1}\left(x+\frac{13-x}{x+1}\right)=42 \text {. }
$$
Three, let $y=\frac{13-x}{x+1}$. The original equation becomes
$x y(x+y)=42$.
$\quad x y+(x+y)=\frac{13 x-x^{2}}{x+1}+\frac{x^{2}+13}{x+1}=13$.
From (1), (2), and Vieta's formulas, we know that $x y$ and $x+y$ are the roots of the equation $z^{2}-13 z+42=0$. Solving this, we get $z_{1}=6, z_{2}=7$.
Thus, we have $\left\{\begin{array}{l}x+y=7, \\ x y=6\end{array}\right.$ or $\left\{\begin{array}{l}x+y=6, \\ x y=7 .\end{array}\right.$
This leads to $x_{1}=1, x_{2}=6, x_{3}=3+\sqrt{2}, x_{4}=3-\sqrt{2}$.