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如果有理数x，y满足|x-1|+（xy-2）2=0．（1）求x，y的值；（2）试求1xy+1(x+1)(y+1)+1(x+2)(y+2)+…+1(x+2009)(y+2009)的值．
题目详情
2
+
+
+…+
的值．
1 1 xy xy
1 1 (x+1)(y+1) (x+1)(y+1)
1 1 (x+2)(y+2) (x+2)(y+2)
1 1 (x+2009)(y+2009) (x+2009)(y+2009)
| 1 |
| xy |
| 1 |
| (x+1)(y+1) |
| 1 |
| (x+2)(y+2) |
| 1 |
| (x+2009)(y+2009) |
| 1 |
| xy |
| 1 |
| (x+1)(y+1) |
| 1 |
| (x+2)(y+2) |
| 1 |
| (x+2009)(y+2009) |
▼优质解答
答案和解析
（1）∵x-1|≥0，（xy-2）22≥0，又|x-1|+（xy-2）22=0
∴|x-1|=0；（xy-2）22=0
∴x=1，y=2；
（2）原式=
+
+
+
+…+
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 12 2 2+
+
+
+…+
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 12×3 2×3 2×3+
+
+…+
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 13×4 3×4 3×4+
+…+
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 14×5 4×5 4×5+…+
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 12010×2011 2010×2011 2010×2011
=
+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 12 2 2+
-
+
-
+
-
+…+
-
+
-
=
．
1 1 12 2 2-
+
-
+
-
+…+
-
+
-
=
．
1 1 13 3 3+
-
+
-
+…+
-
+
-
=
．
1 1 13 3 3-
+
-
+…+
-
+
-
=
．
1 1 14 4 4+
-
+…+
-
+
-
=
．
1 1 14 4 4-
+…+
-
+
-
=
．
1 1 15 5 5+…+
-
+
-
=
．
1 1 12009 2009 2009-
+
-
=
．
1 1 12010 2010 2010+
-
=
．
1 1 12010 2010 2010-
=
．
1 1 12011 2011 2011=
．
2010 2010 20102011 2011 2011．
∴|x-1|=0；（xy-2）22=0
∴x=1，y=2；
（2）原式=
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 2010×2011 |
=
| 1 |
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| 1 |
| 2 |
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| 1 |
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| 1 |
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| 2009 |
| 1 |
| 2010 |
| 1 |
| 2010 |
| 1 |
| 2011 |
| 2010 |
| 2011 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 2010×2011 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
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| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 2009 |
| 1 |
| 2010 |
| 1 |
| 2010 |
| 1 |
| 2011 |
| 2010 |
| 2011 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 2010×2011 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
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| 2010 |
| 1 |
| 2010 |
| 1 |
| 2011 |
| 2010 |
| 2011 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| 2010×2011 |
=
| 1 |
| 2 |
| 1 |
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| 1 |
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| 1 |
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| 1 |
| 2010 |
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| 2010 |
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| 2011 |
| 2010 |
| 2011 |
| 1 |
| 4×5 |
| 1 |
| 2010×2011 |
=
| 1 |
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| 1 |
| 2 |
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| 2010 |
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| 2010 |
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| 2011 |
| 1 |
| 2010×2011 |
=
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