Question

Gia su a,b,c la cacso thoa man a+b+c=259 va \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=15\). Khi do gia tri cua bieu thuc \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)bang

Answer 1

Ta có 

\(Q+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)

\(=259.15\)

\(\Rightarrow Q=259.15-3=3885\)