定义函数序列: $f_{1}(x)=f(x)=\frac{x}{1-x}, f_{2}(x)=f\left(f_{1}(x)\right), f_{3}(x)=f\left(f_{2}(x)\right)$,
$\cdots, f_{n}(x)=f\left(f_{n-1}(x)\right)$, 则函数 $y=f_{2017}(x)$ 的图象与曲线 $y=\frac{1}{x-2017}$ 的交点坐标为 $(\quad)$
由题意 $f_{1}(x)=f(x)=rac{x}{1-x}, f_{2}(x)=f\left(f_{1}(x)
ight)=rac{rac{x}{1-x}}{1-rac{x}{1-x}}=rac{x}{1-2 x}$,$f_{3}(x)=f\left(f_{2}(x)
ight)=rac{rac{x}{1-2 x}}{1-rac{x}{1-2 x}}=rac{x}{1-3 x}$,$f_{n}(x)=f\left(f_{n-1}(x)
ight)=rac{x}{1-n x}$,$	herefore f_{2017}(x)=rac{x}{1-2017 x}$由 $\left\{egin{array}{l}y=rac{1}{x-2017} \ y=rac{x}{1-2017 x}\end{array}
ight.$ ，得 $\left\{egin{array}{l}x=1 \ y=-rac{1}{2016}\end{array}
ight.$ 或 $\left\{egin{array}{l}x=-1 \ y=-rac{1}{2018}\end{array}
ight.$,由 $f_{1}(x)=f(x)=rac{x}{1-x}$ 中 $x 
eq 1$ 得:函数 $y=f_{2017}(x)$ 的图象与曲线 $y=rac{1}{x-2017}$ 的交点坐标为 $\left(-1,-rac{1}{2018}
ight)$,故选 A.