Matematyka.pl

Witam,
rozwiązanie 2 moje:
\frac{a}{a-c} + \frac{b}{b-c} = \frac{2c-b}{2c-b-c} + \frac{b}{b-c} = \frac{2c-b}{c-b} + \frac{b}{b-c} = \frac{2c-b}{c-b} + \frac{b}{-1(-b+c)} = \frac{2c-b}{c-b} + \frac{-b}{c-b} = \frac{2c-b}{c-b} + \frac{-b}{c-b} = \frac{2c-2b}{c-b} = \frac{2(c-b)}{(c-b)} = \frac{2}{1 ...