!!!!!15б!!!срочно!!!!! 1)Докажите тождество:

Решите задачу:

$\frac{2}{(b-c)(c-a)} - \frac{2}{(a-b)(c-b)} + \frac{2}{(a-c)(b-a)} =0 \\ \\ \frac{2}{(b-c)*-(a-c)} - \frac{2}{(a-b)*-(b-c)} + \frac{2}{(a-c)*-(a-b)}= 0\\ \\ - \frac{2}{(b-c)(a-c)} + \frac{2}{(a-b)(b-c)} - \frac{2}{(a-c)(a-b)} =0 \\ \\ \frac{-2(a-b)+2(a-c) -2(b-c)}{(a-b)(b-c)(a-c)} =0 \\ \\ \frac{-2a+2b+2a-2c-2b+2c}{(a-b)(b-c)(a-c)} =0 \\ \\ \frac{(-2a+2a)+(2b-2b) + (-2c+2c)}{(a-b)(b-c)(a-c)} =0 \\ \\ \frac{0}{(a-b)(b-c)(a-c)} =0 \\ \\ 0=0$

$\frac{1}{x} + \frac{1}{a} =b \\ \\ \frac{1}{x} =b - \frac{1}{a} \\ \\ \frac{1}{x} = \frac{ab-1}{a} \\ \\ x * (ab-1) = a*1 \\ \\ x * (ab-1) = a \\ \\ x= \frac{a}{ab-1} \\ \\$