|a-2|+（ab-c²）=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2006)(b+2006)的值

|a-2|+（ab-c²）=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2006)(b+2006)的值

|a-2|+（ab-2）平方=0,
a-2=0,ab=2
所以
a=2,b=1
从而
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2006)(b+2006)
=1/1×2＋1/2×3＋1/3×4＋.+1/2007×2008
=1-1/2+1/2-1/3+1/3-1/4+.+1/2007-1/2008
=1-1/2008
=2007/2008