6. Let $x>0, y>0, z>0$, satisfying $x+y=xy, x+y+z=xyz$, then the range of $z$ is ( )
A: $(0, \sqrt{3}]$
B: $(1, \sqrt{3}]$
C: $\quad\left(0, \frac{4}{3}\right]$
D: $\quad\left(1, \frac{4}{3}\right]$
Answer D.
Analysis Given $y=\frac{x}{x-1}$, so $z=\frac{x+y}{x y-1}=\frac{x+\frac{x}{x-1}}{\frac{x^{2}}{x-1}-1}=\frac{x^{2}}{x^{2}-x+1}=\frac{1}{1-\frac{1}{x}+\frac{1}{x^{2}}}$. Since $x>1$, i.e., $0<\frac{1}{x}<1$, we have $\frac{3}{4} \leq 1-\frac{1}{x}+\frac{1}{x^{2}}<1$, thus $1<z \leq \frac{4}{3}$.