Question

СРОЧНО. Даю 63 балла.

Answer 1

Ответ:   0 .

Объяснение:

$a=0,7\ \ ,\ \ b=-1,2\ \ ,\ \ c=4,3\\\\\dfrac{1}{(a-b)(a-c)}-\dfrac{1}{(a-b)(b-c)}+\dfrac{1}{(c-a)(c-b)}=\\\\\\=\dfrac{1}{(a-b)(a-c)}-\dfrac{1}{(a-b)(b-c)}+\dfrac{1}{(a-c)(b-c)}=\\\\\\=\dfrac{(b-c)-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}=\dfrac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}=\\\\\\=\dfrac{0}{(a-b)(a-c)(b-c)}=0$