2 1) x²+20 = x-3 - 6 x²-4 x+2 x²-x 2) 5 - x-5 - 9 = 0 x²-4x x+2+x x²-49

1) $$ \frac{x^2+20}{x^2-4} = \frac{x-3}{x+2} - \frac{6}{x^2-x} $$

2) $$ \frac{5}{x^2-4x} - \frac{x-5}{x+2+x} - \frac{9}{x^2-49} = 0 $$

Ответ: 1) $$ \frac{x^2+20}{x^2-4} = \frac{x-3}{x+2} - \frac{6}{x^2-x}$$, 2) $$ \frac{5}{x^2-4x} - \frac{x-5}{x+2+x} - \frac{9}{x^2-49} = 0$$