Question

问数学题

Answer 1

∵│ab-2│+(b-1)²=0
又│ab-2│≥0,(b-1)²≥0
∴ab=2,b=1
∴a=2,b=1
∴原式=1/(1×2)+1/(2×3)+..+1/(2008×2009)
=1-1/2+1/2-1/3+..+1/2008-1/2009
=1-1/2009
=2008/2009

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Answer 2

即ab-2=0,b-1=0
所以b=1,a=2/b=1
所以原式=1/1×2+1/2×3+……+1/2008×2009
=1-1/2+1/2-1/3+……+1/2008-1/2009
=1-1/2009
=2008/2009

Answer 3

因为ab-2=0 (b-1)=0,所以ab=2，b=1. 则a=2
.即求1/2+1/（2*3）+1/(3*4)+1/(4*5).......+1/(2009*2008)
=1/2+1/2-1/3+1/3-1/4+1/4-1/5.........+1/2008-1/2009
=1/2+1/2-1/2009
=2008/2009

Answer 4

由|ab-2|+(b-1)^2=0得知：b=1 a=2 (根据非负性）
∴原式=1/2+1/6+......+1/(2008*2009)=(1-1/2)+(1/2-1/3)+......+(1/2008-1/2009)=1/1-1/2009=2008/2009
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