Question

10 задание срочно прошу!!!

Answer 1

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