问题
填空题
已知
|
答案
根据非负数性质可知a-1=0且ab-2=0
解得a=1 b=2
则原式=
+1 1×2
+…+1 2×3 1 2009×2010
裂项得1-
+1 2
-1 2
+1 3
-1 3
+… +1 4
-1 2009
=1-1 2010
=1 2010
;2009 2010
故答案为2009 2010
已知
|
根据非负数性质可知a-1=0且ab-2=0
解得a=1 b=2
则原式=
+1 1×2
+…+1 2×3 1 2009×2010
裂项得1-
+1 2
-1 2
+1 3
-1 3
+… +1 4
-1 2009
=1-1 2010
=1 2010
;2009 2010
故答案为2009 2010