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计算:(1/2009-1/2010)+(1/2010-1/2011)+(1/2011-1/2009)= 注:我打的括号等
题目内容:
计算:(1/2009-1/2010)+(1/2010-1/2011)+(1/2011-1/2009)= 注:我打的括号等于绝对值.
计算:
(1/2-1)+(1/3-1/2)+……+(1/99-1/98)+(1/100-1/99)=
注:我打的括号等于绝对值.优质解答
1/2009>1/2010>1/2011
(1/2009-1/2010)+(1/2010-1/2011)+(1/2011-1/2009)
=1/2009-1/2010+1/2010-1/2011+1/2009-1/2011
=2/2009-2/2011=(2*2011-2*2009)/(2009*2011)
=2*(2011-2009)/(2010-1)*(2010+1)
=4/(2010^2-1)
=4/4040099
利用相同方法去绝对值:(注:此括号不为绝对值)
(1/2-1)+(1/3-1/2)+……+(1/99-1/98)+(1/100-1/99)
=(1-1/2)+(1/2-1/3)+……+(1/98-1/99)+(1/99-1/100)
=1-1/100(当中全抵消了)
=99/100
(绝对值的几何意义是数轴上的点到原点的距离,所以一定是正数,只需比较两数大小即可)
计算:
(1/2-1)+(1/3-1/2)+……+(1/99-1/98)+(1/100-1/99)=
注:我打的括号等于绝对值.
优质解答
(1/2009-1/2010)+(1/2010-1/2011)+(1/2011-1/2009)
=1/2009-1/2010+1/2010-1/2011+1/2009-1/2011
=2/2009-2/2011=(2*2011-2*2009)/(2009*2011)
=2*(2011-2009)/(2010-1)*(2010+1)
=4/(2010^2-1)
=4/4040099
利用相同方法去绝对值:(注:此括号不为绝对值)
(1/2-1)+(1/3-1/2)+……+(1/99-1/98)+(1/100-1/99)
=(1-1/2)+(1/2-1/3)+……+(1/98-1/99)+(1/99-1/100)
=1-1/100(当中全抵消了)
=99/100
(绝对值的几何意义是数轴上的点到原点的距离,所以一定是正数,只需比较两数大小即可)
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