Question

[Nguyễn Thị Diễm Quỳnh]

1.Nếu \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\) thì gtbt \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)

Answer 1

[ngonhuminh]

Bài này trên diễn đàn có nhiều thực chưa có bài thực sự đúng

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

đk: \(\left\{{}\begin{matrix}x+y\ne0\\x+z\ne0\\y+z\ne0\end{matrix}\right.\) Nếu x+y+z=0\(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)(*)

Thay (*) vào (1)

\(\dfrac{x}{-x}+\dfrac{y}{-y}+\dfrac{z}{-z}=-3\) kết luận: \(x+y+z\ne0\)

Nhân 2 vế (1) với x+y+z khác 0 ta có\(\left(1\right)\Leftrightarrow\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=\left(x+y+z\right)\)

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

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

Answer 2

[ngonhuminh]

Vẫn lỗi:

\(.....\\ \left(x+z\right)\dfrac{x}{y+z}+\left(z+x\right)\dfrac{y}{z+x}+\left(x+y\right)\dfrac{z}{x+y}\)

....