问题
选择题
已知
|
答案
∵
=0,(x2-1)2+||xy|-2| (x+1)(y+2)
∴x2-1=0,|xy|-2=0,x+1≠0,y+2≠0,
∴x=1,y=2,
∴原式=
+1 1×2
+…+1 2×3 1 2002×2003
=1-
+1 2
-1 2
+…+1 3
-1 2002 1 2003
=1-1 2003
=
.2002 2003
故选C.
已知
|
∵
=0,(x2-1)2+||xy|-2| (x+1)(y+2)
∴x2-1=0,|xy|-2=0,x+1≠0,y+2≠0,
∴x=1,y=2,
∴原式=
+1 1×2
+…+1 2×3 1 2002×2003
=1-
+1 2
-1 2
+…+1 3
-1 2002 1 2003
=1-1 2003
=
.2002 2003
故选C.