Question

Решите дробное уравнение под номером 4 ппжжпжжп пжжжжжж

Answer 1

№4.
$\frac{x+9}{x-3} - \frac{6}{x^2 - 9} : \frac{6}{(x+3)^2} - \frac{3x-3}{x-3} = \\ \\ =\frac{x+9}{x-3} - \frac{6}{(x-3)(x+3)} * \frac{(x+3)(x+3)}{6} - \frac{3x-3}{x-3} = \\ \\ = \frac{x+9}{x-3} - \frac{1*(x+3)*1}{(x-3)*1*1} - \frac{3x-3}{x-3} = \\ \\ = \frac{x+9-(x+3)-(3x-3)}{x-3} = \frac{x+9-x-3-3x+3}{x-3} = \\ \\ = \frac{-3x+9}{x-3} = \frac{-3(x-3)}{x-3} = \frac{-3}{1} = - 3$