Решите пожалуйста номер 18

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

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