Question

[Bắp Ngô]

Cho\(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2009\)

Tính \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)

Answer 1

[Kiệt Nguyễn]

Xét hiệu :

\(\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\right)-\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)\)

\(=\frac{x^2-y^2}{x+y}+\frac{y^2-z^2}{y+z}+\frac{z^2-x^2}{z+x}\)

\(=\frac{\left(x+y\right)\left(x-y\right)}{x+y}+\frac{\left(y+z\right)\left(y-z\right)}{y+z}+\frac{\left(z+x\right)\left(z-x\right)}{z+x}\)

\(=x-y+y-z+z-x=0\)

Vậy \(\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\right)=\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)\)

hay \(\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)=2009\)