已知根号m-1+根号n-2=0,则1⼀mn+1⼀(m+1)(n+1)+1⼀(m+2)(n+2)+...1⼀（m+2010）（n+2010）的值。

算术平方根恒非负，两算术平方根之和=0，两算术平方根都=0
m-1=0 m=1
n-2=0 n=2
n=m+1
1/(mn)+1/[m+1)(n+1)]+...+1/[(m+2010)(n+2010)]
=1/[m(m+1)]+1/[(m+1)(m+2)]+...+1/[(m+2010)(m+2011)]
=1/m -1/(m+1)+1/(m+1)-1/(m+2)+...+1/(m+2010)-1/(m+2011)
=1/m -1/(m+2011)
=1/1-1/(1+2011)
=1-1/2012
=2011/2012

根号m-1>=0 ; 根号n-2>= 0 ;根号m-1+根号n-2=0 因此m=1,n=2
1/(m+k)(n+k) = 1/(k+1)(k+2) = [(k+2) -(k+1)] / [(k+1)(k+2)] = 1/(k+1)-1/(k+2)
所以
1/mn+1/(m+1)(n+1)+1/(m+2)(n+2)+...1/（m+2010）（n+2010）
= 1-1/2+1/2-1/3+....+1/2011-1/2012
=1-1/2012
=2011/2012

正在算，过几天回答