Giải:
Ta có: \(x-y-z=0\Rightarrow\left\{{}\begin{matrix}x-y=z\\x-z=y\\y+z=x\end{matrix}\right.\)
\(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{y+z}{z}\)
\(=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
Vậy B = -1