Question

Решите пожалуйста. Даю 30 баллов.

Answer 1

Ответ:

$\displaystyle \frac{1}{(a-b)(a-c)}-\frac{1}{(a-b)(b-c)}+\frac{1}{(c-a)(c-b)}=\\\\\\=\frac{1}{(a-b)(a-c)}-\frac{1}{(a-b)(b-c)}+\frac{1}{(a-c)(b-c)}=\\\\\\=\frac{(b-c)-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}=\frac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}=\frac{0}{(a-b)(a-c)(b-c)}=\\\\\\=0$