Có $\sqrt{1+\frac{1}{x^2}+\frac1{(x+1)^2}}=1+\frac 1x-\frac 1{x+1}$Nên $A=1+\frac 12-\frac13+1+\frac 13-\frac 14...+1+\frac 1{9999}-\frac 1{2000}$$=1998+\frac 12-\frac 1{2000}$
Ta có $\sqrt{1+\frac{1}{x^2}+\frac1{(x+1)^2}}=1+\frac 1x-\frac 1{x+1}$Nên $A=1+\frac 12-\frac13+1+\frac 13-\frac 14...+1+\frac 1{9999}-\frac 1{2000}$$=1998+\frac 12-\frac 1{2000}$