Question

[Đinh Tuấn Việt]

Cho \(\dfrac{x}{y+z}+\dfrac{y}{z+z}+\dfrac{z}{x+y}=1\)

Tính \(A=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)

Answer 1

[Mai Thành Đạt]

từ giả thiết ta có \(\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{x\left(y+z\right)}{y+z}+\dfrac{y^2}{y+z}+\dfrac{x\left(y+z\right)}{y+z}+\dfrac{z^2}{x+y}+\dfrac{z\left(x+y\right)}{x+y}=x+y+z\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z=x+y+z\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)

Answer 2

[Dennis]

sai giả thiết