Đáp án:
$\\$
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`
