已知Iab-2I+Ia-2I=0，求1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2006)(b+2006)的值
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由|ab-2|+|a-2|=0得出 ab-2 = 0；a-2 = 0 知 a=2,b=1;
故1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2006)(b+2006)
=1/(2*1) + 1/(3*2) + 1/(4*3) +......+1/(2008*2007)
=(1-1/2) + (1/2 - 1/3) + (1/3-1/4) +...+(1/2007-1/2008)
=1-1/2008
=2007/2008
故1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2006)(b+2006)
=1/(2*1) + 1/(3*2) + 1/(4*3) +......+1/(2008*2007)
=(1-1/2) + (1/2 - 1/3) + (1/3-1/4) +...+(1/2007-1/2008)
=1-1/2008
=2007/2008
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