6.146. $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq-2, \\ x \neq-3, \\ x \neq-4 .\end{array}\right.$

$$
\begin{aligned}
& \text { From the condition } \frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}-1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(x-1)(x-4)(x-2)(x-3)-(x+1)(x+4)(x+2)(x+3)}{(x+1)(x+4)(x+2)(x+3)}=0
\end{aligned}
$$

We have $(x-1)(x-4)(x-2)(x-3)-(x+1)(x+4)(x+2)(x+3)=0$ for $x \neq-1$, $x \neq-2, x \neq-3, x \neq-4 ;\left(x^{2}-5 x+4\right)\left(x^{2}-5 x+6\right)-\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)=0$, $\left(x^{2}-5 x\right)^{2}+10\left(x^{2}-5 x\right)+24-\left(x^{2}+5 x\right)^{2}-10\left(x^{2}+5 x\right)-24=0 \Leftrightarrow$ $\Leftrightarrow-20 x\left(x^{2}+5\right)=0$, from which $x_{1}=0, x^{2}+5 \neq 0$.

Answer: $x=0$.