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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 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
5 |
1 |
2009 |
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2010 |
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2010 |
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2011 |
2010 |
2011 |
1 |
2 |
1 |
2×3 |
1 |
3×4 |
1 |
4×5 |
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2010×2011 |
=
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
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 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
5 |
1 |
2009 |
1 |
2010 |
1 |
2010 |
1 |
2011 |
2010 |
2011 |
1 |
3×4 |
1 |
4×5 |
1 |
2010×2011 |
=
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
5 |
1 |
2009 |
1 |
2010 |
1 |
2010 |
1 |
2011 |
2010 |
2011 |
1 |
4×5 |
1 |
2010×2011 |
=
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
5 |
1 |
2009 |
1 |
2010 |
1 |
2010 |
1 |
2011 |
2010 |
2011 |
1 |
2010×2011 |
=
1 |
2 |
1 |
2 |
1 |
3 |
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3 |
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1 |
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1 |
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2009 |
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