Question

Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\). Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

Answer 1

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 ... @@

Answer 2

Cộng 3 zô :v

Wait for tui five phút

Answer 3

P/s : Nếu cần c/m x+y+x khác không thùy công 3 zô :3