a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b_数学解答

a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值要用的
a,b满足根号a-1+根号b-2=0 求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)的值

人气：357 ℃　时间：2020-02-05 22:07:36

解答

√a-1+√b-2=0a=1,b=21/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)=1/2+1/2*3+1/3*4+1/4*5+...+1/2009*2010=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010=1-1/2010=2009/2010