Solve for $x \in \mathbb{R}$, $\frac{{(x + 2)(x + 3)(x + 4)(x + 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}} = 1$

We have, $\frac{{({x^2} + 5x + 6)({x^2} + 9x + 20)}}{{({x^2} - 5x + 6)({x^2} - 9x + 20)}} = 1$

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} - 5x + 6}} = \frac{{{x^2} - 9x + 20}}{{{x^2} + 5x + 6}} = \frac{{({x^2} + 9x + 20) - ({x^2} - 9x + 20)}}{{({x^2} - 5x + 6) - ({x^2} + 5x + 6)}} = \frac{{18x}}{{ - 10x}} =  - \frac{{9}}{{5}}$
(assuming $x \ne 0$)

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} - 5x + 6}} - 1 =  - \frac{9}{5} - 1$

$ \Rightarrow \frac{{14x + 14}}{{{x^2} - 5x + 6}} =  - \frac{{14}}{5}$

$ \Rightarrow \frac{{x + 1}}{{{x^2} - 5x + 6}} =  - \frac{1}{5}$

$ \Rightarrow {x^2} - 5x + 6 + 5x + 5 = 0$ Or ${x^2} + 11 = 0$

So, no real solution if $x \neq 0$.

If x = 0, LHS of the original equation = $\frac{{2 \times 3 \times 4 \times 5}}{{( - 2) \times ( - 3) \times ( - 4) \times ( - 5)}}$ = 1 = RHS

Thus, x = 0 is the only real solution.