`Answer:`
Mình sửa đề lại thành: \(F=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)
Theo đề ra, ta có: \(-x+y-z=0\Rightarrow\hept{\begin{cases}y=x+z\\x=y-z\\y-x=z\end{cases}}\left(\text{*}\right)\)
\(F=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)=\left(\frac{z}{z}+\frac{x}{z}\right)\left(\frac{x}{x}-\frac{y}{x}\right)\left(\frac{y}{y}-\frac{z}{y}\right)=\frac{z+x}{z}.\frac{-\left(y-x\right)}{x}.\frac{y-z}{y}\)
Thay (*) vào `F:` \(F=\frac{y}{z}.\frac{-z}{x}.\frac{x}{y}=-1\)