Докажите, что верно равенство 1/((a-b)(a-c))+1/((b-a)(b-c))-1/((c-a)(b-c))=0.

Ответ:

\[\frac{1}{(a - b)(a - c)} + \frac{1}{(b - a)(b - c)} - \frac{1}{(c - a)(b - c)} = 0\]

\[\frac{1^{\backslash b - c}}{(a - b)(a - c)} - \frac{1^{\backslash a - c}}{(a - b)(b - c)} + \frac{1^{\backslash a - b}}{(a - c)(b - c)} = 0\]

\[\frac{b - c - a + c + a - b}{(a - b)(a - c)(b - c)} = 0\ \ \]

\[0 = 0\]

\[ч.т.д.\]