Срочно нужна помощь помогите в течение 15 минут

Решите задачу:

$\frac{ \frac{1}{a}- \frac{1}{b-c} }{\frac{1}{a}+\frac{1}{b-c} } \cdot(1- \frac{b^2+c^2-a^2}{2bc}): \frac{b-a-c}{abc}, npu \ a=1,2; \ b=0,5; \ c=1,3$

$1. \ \frac{1}{a}- \frac{1}{b-c} = \frac{b-c-a}{a(b-c)}\\\\\\ 2. \ \frac{1}{a}+ \frac{1}{b-c} =\frac{b-c+a}{a(b-c)}\\\\\\ 3. \ \frac{b-c-a}{a(b-c)}\cdot \frac{a(b-c)}{b-c+a}=\frac{b-c-a}{b-c+a}= \frac{-a+b-c}{a+b-c} \\\\\\ 4. \ (1- \frac{b^2+c^2-a^2}{2bc})= \frac{2bc-b^2-c^2+a^2}{2bc}= \frac{-(b^2-2bc+c^2)+a^2}{2bc}=\\\\ = \frac{a^2-(b-c)^2}{2bc}= \frac{(a-b+c)(a+b-c)}{2bc} \\\\\\$

$5. \ \frac{-a+b-c}{a+b-c} \cdot\frac{(a-b+c)(a+b-c)}{2bc} = \frac{(a-b+c)(-a+b-c)}{2bc}= \frac{-(a-b+c)(a-b+c)}{2bc}=\\\\ =- \frac{(a-b+c)^2}{2bc} \\\\\\ 6. \ - \frac{(a-b+c)^2}{2bc} \cdot \frac{abc}{b-a-c}=- \frac{a(a-b+c)^2}{2(b-a-c)}= \frac{a(a-b+c)^2}{2(a-b+c)}= \frac{a(a-b+c)}{2} \\\\\\ \frac{a(a-b+c)}{2}, \ npu \ a=1,2; \ b=0,5; \ c=1,3\\\\ \frac{1,2\cdot(1,2-0,5+1,3)}{2}= \frac{1,2\cdot2}{2} =1,2$