已知4a²-4ab+2b²-4b+4=0,试求1/ab+1/（a+1)(b+1)+.1/（a+2009)(b+2009)的值?

已知4a²-4ab+2b²-4b+4=0,试求1/ab+1/（a+1)(b+1)+.1/（a+2009)(b+2009)的值?

4a²-4ab+2b²-4b+4=0
4a²-4ab+b²+b²-4b+4=0
(2a-b)²+(b-2)²=0
由非负性
2a-b=0
b-2=0
解得
a=1
b=2
则
原式
=1/(1*2)+1/(2*3)+...+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+...+(1/2010-1/2011)
=1-1/2011
=2010/2011

4a²-4ab+2b²-4b+4=0
(2a-b)^2+(b-2)^2=0
2a-b=0,b=2
a=1
1/ab+1/（a+1)(b+1)+.....1/（a+2009)(b+2009)
=1/2+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011