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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: `(1/2+1/3+…+1/(2009)+1/(2010))/((2009)/1+(2008)/2+…+2/(2008)+1/(2009))`

Toán Lớp 7: (1/2+1/3+…+1/(2009)+1/(2010))/((2009)/1+(2008)/2+…+2/(2008)+1/(2009))

Comments ( 2 )

  1. Giải đáp:
    $\\$
    Đặt A= (1/2 + 1/3 + … + 1/2009 + 1/2010)/(2009/1 + 2008/2 + … + 2/2008 + 1/2009)
    Đặt B = 2009/1 + 2008/2 + … + 2/2008 + 1/2009
    -> B = 2009 + 2008/2 + … + 2/2008 + 1/2009
    -> B = 1 + (2008/2 + 1) + … + (2/2008 + 1) + (1/2009+1)
    -> B = 1 + (2008/2 + 2/2) + … + (2/2008 + 2008/2008) + (1/2009 + 2009/2009)
    -> B = 2010/2010 + 2010/2 + … + 2010/2008 + 2010/2009
    -> B = 2010 (1/2010 + 1/2009 + 1/2008 + … + 1/2)
    Thay vào biểu thức A ta được :
    -> A = (1/2 + 1/3 + … + 1/2009 + 1/2010)/(2010 (1/2 + 1/3 + … + 1/2009 + 1/2010) )
    -> A = 1/2010
    Vậy A=1/2010
     

  2. Giải đáp:(1/2+1/3+….+1/2009+1/2010)/(2009/1+2008/2+……+2/2008+1/2009)=1/2010.
     
    Lời giải và giải thích chi tiết:
    Tách 2009=\underbrace{1+1+…+1}_{\text{2009 số hạng}}
    (1/2+1/3+….+1/2009+1/2010)/(2009/1+2008/2+……+2/2008+1/2009)
    =(1/2+1/3+….+1/2009+1/2010)/(1/2009+2/2008+….+2008/2+1.2009)
    =(1/2+1/3+….+1/2009+1/2010)/(1+1/2009+1+2/2008+…..+1+2008/2)
    =(1/2+1/3+….+1/2009+1/2010)/(2010/2009+2010/2008+…..+2010/2)
    =(1/2+1/3+….+1/2009+1/2010)/(2010(1/2+1/3+….+1/2009+1/2010))
    =1/2010.

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222-9+11+12:2*14+14 = ? ( )