Question

Помогите пожалуйста решить

Answer 1

$\displaystyle\frac{\displaystyle\frac{1}{a}+\frac{1}{bc}}{\displaystyle\frac{1}{a}-\frac{1}{b+c}}*(1+\frac{b^2+c^2-a^2}{2bc})*(a+b+c)^{-2}=\\=\frac{\displaystyle\frac{bc+a}{abc}}{\displaystyle\frac{b+c-a}{a(b+c)}}*(\frac{2bc+b^2+c^2-a^2}{2bc})*(a+b+c)^{-2}=\\=\frac{\displaystyle\frac{bc+a}{abc}}{\displaystyle\frac{b+c-a}{a(b+c)}}*(\frac{(b+c)^2-a^2}{2bc})*\frac{1}{(a+b+c)^2}=$

$\displaystyle\frac{(bc+a)(a(b+c))}{abc(b+c-a)}*(\frac{(b+c)^2-a^2}{2bc})*\frac{1}{(a+b+c)^2}=\\=\frac{(bc+a)(a(b+c))}{abc(b+c-a)}*(\frac{(b+c-a)(b+c+a)}{2bc})*\frac{1}{(a+b+c)^2}=\\=\frac{(bc+a)(b+c)}{2b^2c^2}*\frac{1}{a+b+c}$