已知：|x-2|+|y-4|=0,求：1/xy+1/(x+2)(y+2)+1/（x+4)(y+4）+……+1/(x+2010)(y+2010).要简便运算.

Question

已知：|x-2|+|y-4|=0,
求：1/xy+1/(x+2)(y+2)+1/（x+4)(y+4）+……+1/(x+2010)(y+2010).要简便运算.

Answer 1

|x-2|+|y-4|=0,绝对值≥0,说明|x-2|=0；|y-4|=0.所以X=2 Y=4
1/xy+1/(x+2)(y+2)+1/（x+4)(y+4）+……+1/(x+2010)(y+2010)
=1/2*4+1/(2+2)(4+2)+1/（2+4)(4+4）+……+1/(2+2010)(4+2010)=1/(2*4)+1/(4*6)+1/(4*6）+……+1/(2012*2014)
=1/2[(1/2-1/4)+(1/4-1/6)+……+(1/2012-1/2014)
]=1/2*(1/2-1/2014)
=503/2014
于22点21分50秒上传过程详细绝对原创!