若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
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若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
若ab-2的绝对值+(b-2)的平方=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
易得b=2;a=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)
=1/2+1/2-1/3+1/3-1/4+······1/2010-1/2011
=1-1/2011
=2010/2011
∵|ab-2|+(b-2)²=0
∴ab=2 b=2
∴a=1 b=2=a+1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
=1/a-1/b+1/(a+1)-1/(b+1)+……+1/(a+2009)-1/(b+2009)
=1/a-1/(a+1)+1/(a+1)-1/(b+2)+...
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∵|ab-2|+(b-2)²=0
∴ab=2 b=2
∴a=1 b=2=a+1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
=1/a-1/b+1/(a+1)-1/(b+1)+……+1/(a+2009)-1/(b+2009)
=1/a-1/(a+1)+1/(a+1)-1/(b+2)+……+1/(a+2009)-1/(a+2010)
=1/a-1/(a+2010)
=1/1-1/2011
=1-1/2011
=2010/2011
收起
因为是绝对值和开放;所以如等于零;则;ab-2=0并且b=2;所以a=1
剩下的那个方程你代入算吧;我真的忘记了。