初一的数学题。要快！！！！

|a|+a=0,|ab|=ab,|c|-c=0
那么a<0 ab>0 c>0 得b<0
|b|-|a+b|-|c-b|+|a-c|
=-b+(a+b)-(c-b)-(a-c)
=b

1/1×2+1/2×3+1/3×4+1/4×5+……+1/2009×2010
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....+(1/2009-1/2010)
=1-1/2010
=2009/2010

1.由已知可得：
a≤0 ab≥0 b≤0 c≥0
∴原式=-b+（a+b)-(c-b)-(a-c)
=b
2.1/1×2+1/2×3+1/3×4+1/4×5+……+1/2009×2010
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....+(1/2009-1/2010)
=1-1/2010
=2009/2010

|a|+a=0,
a<0

|ab|=ab,
b<0

|c|-c=0
c>0

|b|-|a+b|-|c-b|+|a-c|
=-b-(-a-b)-(c-b)+(c-a)
=-b+a+b-c+b+c-a
=b

1/1×2+1/2×3+1/3×4+1/4×5+……+1/2009×2010.
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2009-1/2010
=1-1/2010
=2009/2010