根号a-1加根号b-2等于0，求1/ab+1/(a+)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)……1/(a+2008)(b+2008)

来源：百度知道 编辑：UC知道 时间：2024/05/06 13:12:04

根号求解是非负数~
两个非负数相加要等于0，只有两个都为0
所以根号a-1=0
根号b-2=0
即a=1，b=2
然后代进表达式，就可以求解了！

表达式分解下
=[1/(b-a)]*[(1/a)-(1/b)]+[1/(b-a)]*[(1/(a+1))-(1/(b+1))]+...
=(1-1/2) + 1/2 *(1/2-1/3)....1/2 * (1/2009 - 1/3000)
=( 1-1/2+1/2-1/3....+1/2009-1/2010)
=(1-1/2010)
= 2009/2010

根号下分别为0 a=1 b=2 原式=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5........+1/2009/2010=2009/2010

√a-1 +√b-2=0
a-1=0 b-2=0
a=1 b=2

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)……1/(a+2008)(b+2008)
=1/1*2 +1/2*3 +1/3*4+1/4*5 ...........1/2009*2010
=1-1/2 +1/2-1/3 +1/3-1/4 +1/4-1/5.......1/2009-1/2010
=1-1/2010
=2009/2010