Question

Cho x + y + z = 2007 và \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=\frac{1}{90}\). Tính tổng : \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\).

Answer 1

ta có: \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=\frac{1}{90}.\)

\(\Rightarrow2007.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)=2007\cdot\frac{1}{90}\)

\(\frac{2007}{x+y}+\frac{2007}{y+z}+\frac{2007}{x+z}=\frac{223}{10}\)

mà x+y+z = 2007

\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}=\frac{223}{10}\)

\(1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{x+z}=\frac{223}{10}\)

\(\Rightarrow\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{x+z}=\frac{223}{10}-3=\frac{193}{10}\)