10 задание срочно прошу!!!

Решите задачу:

$\frac{6-x}{1-x^2} - \frac{x+3}{x(1-x)} = \frac{x+5}{x(1+x)}\; ,\; \; x\ne 0\; ,\; x\ne \pm 1 \\\\ \frac{6-x}{(1-x)(1+x)} - \frac{x+3}{x(1-x)} - \frac{x+5}{x(1+x)} =0\\\\ \frac{x(6-x)-(x+3)(1+x)-(x+5)(1-x)}{x(1-x)(1+x)} =0\\\\ \frac{6x-x^2-(x^2+4x+3)-(-x^2-4x+5)}{x*1-x)(1+x)} =0\\\\ \frac{-x^2+6x-8}{x(1-x)(1+x)} =0\; ,\; \; x^2-6x+8=0\\\\x_1=4,\; \; x_2=2\; \; (teorema\; Vieta)$