Lời giải:
Từ $a+b+c=2; \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=2,5$
$\Rightarrow (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=5$
\(\Leftrightarrow \frac{a}{a+b}+\frac{a}{a+c}+\frac{a}{b+c}+\frac{b}{a+b}+\frac{b}{a+c}+\frac{b}{b+c}+\frac{c}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}=5\)
\(\Leftrightarrow \frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=5\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\)
Khi đó:
\(A=\frac{a-(b+c)}{b+c}+\frac{b-(c+a)}{c+a}+\frac{c-(a+b)}{a+b}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-3\)
\(=2-3=-1\)
Vậy $A=-1$
