七年级数学题若X-1的绝对值+(XY-2)平方=0,求1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)的值

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七年级数学题若X-1的绝对值+(XY-2)平方=0,求1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)的值

七年级数学题若X-1的绝对值+(XY-2)平方=0,求1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)的值
七年级数学题若X-1的绝对值+(XY-2)平方=0,求1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)的值

七年级数学题若X-1的绝对值+(XY-2)平方=0,求1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)的值
因为绝对值和平方的结果都不小于0,因此只有两部分都为0时才成立
X-1=0,X=1
XY-2=0,XY=2,Y=2
1/(1×2)=1/1-1/2,1/(2×3)=1/2-1/3,1/(3×4)=1/3-1/4...
所以
原式=1/(1×2)+1/(2×3)+1/(3×4)+...+1/(2012×2013)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2012-1/2013
=1-1/2013
=2012/2013

由题意可得:X-1=XY-2=0,所以X=1,Y=2.因此
1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+……+1/(X+2011)(Y+2011)
=1-1/2+(1/2-1/3)+(1/3-1/4)+……+(1/2012-1/2013)
=1-1/2013
=2012/2013

由非负和等于0,可得x=1,y=2
则1/XY+1/(X+1)(Y+1)+1/(X+2)(Y+2)+----+1/(X+2011)(Y+2011)
=1/2+1/2*3+........+1/2012*2013
=1-1/2+1/2-1/3+...........+1/2012-1/2013
=1-1/2013
=2012/2013

由于 |x-1|>=0 且 (xy-2)^2>=0,而二者之和又等于0,所以有
|x-1|=0 且 (xy-2)^2=0
所以 x=1 y=2
所以 1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+----+1/(x+2011)(y+2011)
= 1/(1*2)+1/(2*3)+1/(3*4)+……+1/...

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由于 |x-1|>=0 且 (xy-2)^2>=0,而二者之和又等于0,所以有
|x-1|=0 且 (xy-2)^2=0
所以 x=1 y=2
所以 1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+----+1/(x+2011)(y+2011)
= 1/(1*2)+1/(2*3)+1/(3*4)+……+1/(2011*2012)+1/(2012*2013)
= 1/1-1/2+1/2-1/3+1/3-1/4+……+1/2011-1/2012+1/2012-1/2013
= 1-1/2013
=2012/2013

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