Question

cho a+b+c=2001

\(\frac{1}{a+b}\)+\(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)=\(\frac{1}{10}\)

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

Answer 1

Ta có :

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

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

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

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

Thay \(a+b+c=2001\)và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10};\)có :

\(A=2001.\frac{1}{10}-3\)

\(=200,1-3\)

\(=197,1\)

Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=197,1\)