如果有理数a,b满足|ab-2|+(1-6)的平方=0试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+
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如果有理数a,b满足|ab-2|+(1-6)的平方=0试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2009)(b+2009)
|ab-2|+(1-b)的平方=0
ab-2=0,1-b=0
a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2009)(b+2009)
=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
ab-2=0,1-b=0
a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2009)(b+2009)
=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
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