Dễ thấy \(x+y+z\ne0\)
Ta có :
\(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}\right)+\left(x+y+z\right)\left(\frac{y}{x+z}\right)+\left(x+y+z\right)\left(\frac{z}{x+y}\right)\)
\(=\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z\)
\(=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z\)
Mà \(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(=\left(x+y+z\right).1=x+y+z\)
=> \(x+y+z=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z\)
=> \(0=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Làm xong thấy bài tui làm hới ... @@
P/s : Nếu cần c/m x+y+x khác không thùy công 3 zô :3
