Question

решите пожалуйста номер 18

Answer 1

 

$frac{1}{(x-y)(y-z)}-frac{1}{(y-z)(x-z)}-frac{1}{(z-x)(y-x)}= \ =frac{1}{(x-y)(y-z)}-frac{1}{(y-z)(x-z)}-frac{1}{(x-z)(x-y)}=frac{(x-z)-(x-y)-(y-z)}{(x-y)(x-z)(y-z)}= \ =frac{x-z-x+y-y+z}{(x-y)(x-z)(y-z)}=0$

 

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