Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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