1. Find at least one root of the equation

$$
\frac{x-36}{x-1}+\frac{x-18}{x-2}+\frac{x-12}{x-3}+\frac{x-9}{x-4}+\frac{x+9}{x+4}+\frac{x+12}{x+3}+\frac{x+18}{x+2}+\frac{x+36}{x+1}=0
$$
1. Solution. Let's regroup the terms as follows:

$\frac{x-36}{x-1}+\frac{x+36}{x+1}+\frac{x-18}{x-2}+\frac{x+18}{x+2}+\frac{x-12}{x-3}+\frac{x+12}{x+3}+\frac{x-9}{x-4}+\frac{x+9}{x+4}=$

$\frac{2 x^{2}-72}{x^{2}-1}+\frac{2 x^{2}-72}{x^{2}-1}+\frac{2 x^{2}-72}{x^{2}-9}+\frac{2 x^{2}-72}{x^{2}-16}$

Now it is clear that the numbers $\pm 6$ are roots of this equation. You can simply guess these roots and perform the verification. Grading criteria. All transformations are provided, after which one or both roots become obvious - 7 points. At least one root is guessed and verified - 7 points. In all other cases - 0 points.