初一的数学题。要快!!!!
1.设a。b。c为实数,且|a|+a=0,|ab|=ab,|c|-c=0,求代数式|b|-|a+b|-|c-b|+|a-c|的值2.计算:1/1×2+1/2×3+1/3×...
1.设a。b。c为实数,且|a|+a=0,|ab|=ab,|c|-c=0,
求代数式|b|-|a+b|-|c-b|+|a-c|的值
2.计算:1/1×2+1/2×3+1/3×4+1/4×5+……+1/2009×2010. 展开
求代数式|b|-|a+b|-|c-b|+|a-c|的值
2.计算:1/1×2+1/2×3+1/3×4+1/4×5+……+1/2009×2010. 展开
3个回答
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|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
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
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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≤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
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|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
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
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