6.003. $\frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$.

6.003. $\frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$.
Solution.

Domain of definition: $x \neq 0, x \neq \frac{-1 \pm \sqrt{21}}{2}$.

Let $\frac{x^{2}+x-5}{x}=z$, then $z+\frac{1}{z}+4=0 \Rightarrow$

$\Rightarrow z^{2}+4 z+3=0, \Rightarrow z_{1}=-3, z_{2}=-1$.

To find $x$, solve the two equations:

$\frac{x^{2}+x-5}{x}=-3$ or $\frac{x^{2}+x-5}{x}=-1$.

Solving each of them, we find:

$x_{1}=-5, x_{2}=1, x_{3}=-1-\sqrt{6}, x_{4}=-1+\sqrt{6}$.

Answer: $x_{1}=-5, x_{2}=1, x_{3,4}=-1 \pm \sqrt{6}$.