Решите уравнение x+3//x^2-x-x-5//x+x^2=x-6//1-x^2 // деление

Решите задачу:

$\frac{x+3}{x^2-x} - \frac{x-5}{x+x^2} = \frac{x-6}{1-x^2} \\\\ \frac{x+3}{x(x-1)} - \frac{x-5}{x(1+x)} = \frac{x-6}{(1-x)(1+x)}\; ,\; \; ODZ:\; x\ne 0\; ,\; x\ne 1\; ,\; x\ne -1\\\\\frac{(x+3)(x+1)-(x-5)(x-1)}{x(x-1)(x+1)} = \frac{x-6}{-(x-1)(x+1)} \\\\\frac{x^2+4x+3-(x^2-6x+5)}{x(x-1)(x+1)}+ \frac{x-6}{(x-1)(x+1)}=0\\\\ \frac{10x-2+x(x-6)}{x(x-1)(x+1)} =0 \\\\ \frac{x^2+4x-2}{x(x-1)(x+1)} =0\\\\x^2+4x-2=0\; ,\; \; \; D/4=4+2=6\\\\x_1= -8-\sqrt8=-8-\sqrt6 \; \; ;\; \; \; x_2=-8+\sqrt{6}$