已知|ab-2|+(b+1)²=0.,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)的值.过程+答案
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![已知|ab-2|+(b+1)²=0.,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)的值.过程+答案](/uploads/image/z/10321814-38-4.jpg?t=%E5%B7%B2%E7%9F%A5%7Cab-2%7C%2B%EF%BC%88b%2B1%29%26sup2%3B%3D0.%2C%E6%B1%821%2Fab%2B1%2F%28a-1%29%28b-1%29%2B1%2F%28a-2%29%28b-2%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28a-2010%EF%BC%89%EF%BC%88b-2010%EF%BC%89%E7%9A%84%E5%80%BC.%E8%BF%87%E7%A8%8B%2B%E7%AD%94%E6%A1%88)
已知|ab-2|+(b+1)²=0.,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)的值.过程+答案
已知|ab-2|+(b+1)²=0.,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)的值.
过程+答案
已知|ab-2|+(b+1)²=0.,求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)的值.过程+答案
因为绝对值和平方都大于等于0,而|ab-2|+(b+1)²=0.,
所以ab-2=0,b+1=0
解得a=-2 ,b=-1
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)
=1/(-2)*(-1)+1/(-2-1)*(-1-1)+1/(-2-2)*(-1-2)……+1/(-2-2010)*(-1-2010)
=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
|ab-2|+(b+1)²=0
所以|ab-2|=0,(b+1)²=0
所以b=-1,a=-2
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)
裂项相消法来做就可以了
|ab-2|+(b+1)²=0
b=-1,a=-2
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)+……+1/(a-2010)(b-2010)=1-1/2+1/2-1/3+1/3-1/4+.......+1/2011-1/2012=1-1/2012=2011/2012