帮我做一道数学题?已知（a-1）的平方+ab-2的绝对值=0求ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1一直这样加到(a+2009)(b+2009)/1

Question

帮我做一道数学题?
已知（a-1）的平方+ab-2的绝对值=0 求ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1一直这样加到
(a+2009)(b+2009)/1

Answer 1

因为(a-1)^2>=0
｜ab-2｜>=0
欲使(a-1)^2+｜ab-2｜=0
所以必有
(a-1)^2=0
｜ab-2｜=0
可得：a=1,b=2
所以1/ab+1/[(a+1)(b+1)]+1/[(a+2)(b+2)]+...+1/[(a+2009)(b+2009)]
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2010*2011)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2010-1/2011)
=1-1/2011
=2010/2011