Question

Cho a+b+c=4034 vµ \(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\) .Tính \(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)

Answer 1

\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)

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

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

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

\(=4034.\dfrac{1}{2}-3=2014\)

Answer 2

Guể?

\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)

\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)

\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)