已知函数 f(x)(x∈R)f\left( x\right) \left( {x \in \mathbf{R}}\right)f(x)(x∈R) 满足 f(−x)=2−f(x)f\left( {-x}\right) = 2 - f\left( x\right)f(−x)=2−f(x) ,若函数 y=x+1xy = \frac{x + 1}{x}y=xx+1 与 y=f(x)y = f\left( x\right)y=f(x) 图像的交点为 (x1,y1),(x2,y2),⋯ ,(xm,ym)\left( {{x}_{1},{y}_{1}}\right) ,\left( {{x}_{2},{y}_{2}}\right) ,\cdots ,\left( {{x}_{m},{y}_{m}}\right)(x1,y1),(x2,y2),⋯,(xm,ym) ,则 ∑i=1m(xi+yi)=\mathop{\sum }\limits_{{i = 1}}^{m}\left( {{x}_{i} + {y}_{i}}\right) =i=1∑m(xi+yi)=A. 0 B. mmm C. 2m{2m}2m D. 4m{4m}4m