Question

Прошу решить это задание!

Answer 1

$frac{2}{x^2-x-12}+frac{6}{x^2+4x+3}=frac{1}{x+3} \\ frac{2}{(x-4)(x+3)}+frac{6}{(x+1)(x+3)}=frac{1}{x+3}; ,; ; ; ODZ:; xne 4; ,xne-3; ,; xne -1\\ frac{2(x+1)+6(x-4)-(x+1)(x-4)}{(x-4)(x+3)(x+1)} =0\\frac{2x+2+6x-24-(x^2-3x-4)}{(x-4)(x+3)(x+1)}=0\\frac{-x^2+11x-18}{(x-4)(x+3)(x+1)} =0\\ frac{x^2-11x+18}{(x-4)(x+3)(x+1)} =0\\x^2-11x+18=0; ; to ; ; x_1=2; ,; x_2=9; (teor,; Vieta)\\Otvet:; ; x=2; ,; ; x=9; .$