B Let 2.00 ... 04 = x and 2.00 ... 02 = y. This means that $x>y.$ We claim that $\frac{x}{(x-1)^2+x}<\frac{y}{(y-1)^2+y}.$ This can be simplified into $\frac{x}{x^2-x+1}<\frac{y}{y^2-y+1}.$ Since $x^2-x+1,y^2-x+1>0,$ we can multiply both sides of the inequality by $(x^2-x+1)(y^2-y+1)$ to obtain $x(y^2-y+1)<y(x^2-x+1).$ This equation can be simplified into $xy^2-yx^2<y-x$ which is equivalent to $xy(y-x)<(y-x).$ After dividing by the negative number $y-x$ from both sides, we obtain the inequality $-xy<-1$ which is obviously true. Hope this helps! Please tell me if there are any errors in my solution.