计算:(1/2009-1/2010)+(1/2010-1/2011)+(1/2011-1/2009)= 注:我打的括号等

2个回答

  • 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)