Question

ab-2的绝对值+（b-1)2次方=0，求ab分之1+（a+1)(b+1）分之1+（a+2)(b+2)分之1+……（a+2010)(b+2010)分之1

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Answer 1

ab-2的绝对值+（b-1)2次方=0
ab=2，b=1
a = 2/b = 2
ab分之1+（a+1)(b+1）分之1+（a+2)(b+2)分之1+……（a+2010)(b+2010)分之1
= 1/(2*1) + 1/(3*2) + 1/(4*3) +......+1/(2012*2011)
= 1-1/2 + 1/2-1/3+1/3-1/4+......1/2011-1/2012
= 1-1/2012
= (2012-1)/2012
= 2011/2012

Answer 2

|ab-2|+（b-1)^2=0
ab=2，b=1
a=2 ,b=1
ab分之1+（a+1)(b+1）分之1+（a+2)(b+2)分之1+……（a+2010)(b+2010)分之1
=1/2+1/6+1/12+……+1/(2012*2011)
=1-1/2 + 1/2-1/3+1/3-1/4+......1/2011-1/2012
= 1-1/2012
= 2011/2012