Question

Cho : x + y + z = 2017 và : \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{1}{672}\)

Tính : \(C=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)

Answer 1

Đặt : \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=M\)

\(\Rightarrow\left(x+y+z\right).M=\frac{1}{672}.2017\)

\(\Rightarrow1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=\frac{2016}{672}+\frac{1}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=3+\frac{1}{672}\)

\(\Rightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{1}{672}\)

Answer 2

Nhân cả 2 vế với \(x+y+z\),ta được:

\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{672}\cdot2017\)

\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=\frac{2017}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{2017}{672}\)

\(\Rightarrow C=\frac{1}{672}\)