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Bài toán `1`

Ta có: `2009^20=(2009^2)^10=4036081^10`

Vì `4  036  081<20  092  009`

Nên `4036081^10<20092009^10` hay `2009^20<20092009^10`

Vậy `2009^20<20092009^10.`$\\$

Bài toán `2`

Ta có: $\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}$

$=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\bigg(\dfrac{2007}{2}+1\bigg)+\bigg(\dfrac{2006}{3}+1\bigg)+...+\bigg(\dfrac{2}{2007}+1\bigg)+\bigg(\dfrac{1}{2008}+1\bigg)}$

$=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}$

$=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009·\bigg(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\bigg)}$

`=1/2009`

Vậy `A/B=1/2009.`