Đáp án:
`A<B`.
Giải thích các bước giải:
`A=(2008^2008+1)/(2008^2009+1`
`=>2008A=(2008.(2008^2008+1))/(2008^2009+1`
`=>2008A=(2008^2009+2008)/(2008^2009+1`
`=>2008A=(2008^2009+1+2007)/(2008^2009+1`
`=>2008A=(2008^2009+1)/(2008^2009+1)+2007/(2008^2009+1`
`=>2008A=1+2007/(2008^2009+1`
`B=(2008^2007+1)/(2008^2008+1`
`=>2008A=(2008.(2008^2007+1))/(2008^2008+1`
`=>2008A=(2008^2008+2008)/(2008^2008+1`
`=>2008A=(2008^2008+1+2007)/(2008^2008+1`
`=>2008A=(2008^2008+1)/(2008^2008+1)+2007/(2008^2008+1`
`=>2008A=1+2007/(2008^2008+1`
`=>2007/(2008^2009+1)<2007/(2008^2008+1`
`=>1+2007/(2008^2009+1)<1+2007/(2008^2008+1`
`=>2008A<2008B`
`=>A<B`
Vậy `A<B`.