Lời giải:
Ta có: \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(Q+3=\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1\)
\(Q+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(Q+3=\frac{259}{a+b}+\frac{259}{b+c}+\frac{259}{a+c}=259\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\)
\(Q+3=259.15=3885\)
\(\Rightarrow Q=3885-3=3882\)