Bài 2 (toán nâng cao)

H

harrypham

Ta có $$\begin{aligned} B= \dfrac{2008}{1}+ \dfrac{2007}{2}+ \dfrac{2006}{3}+ \cdots + \dfrac{1}{2008} & = \dfrac{2009-1}{1}+ \dfrac{2009-2}{2}+ \dfrac{2009-3}{3}+ \cdots + \dfrac{2009-2008}{2008} \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008} \right) + 2009 - 2008 \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008} \right) +1 \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008}+ \dfrac{1}{2009} \right) \end{aligned}$$
Do đó $\dfrac{A}{B}= \boxed{ \dfrac{1}{2009}}$.
 
K

kieuphuong232

harrypham said:
Ta có $$\begin{aligned} B= \dfrac{2008}{1}+ \dfrac{2007}{2}+ \dfrac{2006}{3}+ \cdots + \dfrac{1}{2008} & = \dfrac{2009-1}{1}+ \dfrac{2009-2}{2}+ \dfrac{2009-3}{3}+ \cdots + \dfrac{2009-2008}{2008} \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008} \right) + 2009 - 2008 \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008} \right) +1 \\ & = 2009 \left( \dfrac{1}{2}+ \dfrac{1}{3}+ \cdots + \dfrac{1}{2008}+ \dfrac{1}{2009} \right) \end{aligned}$$
Do đó $\dfrac{A}{B}= \boxed{ \dfrac{1}{2009}}$.
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Mình có 1 cách giải khác:
Ta có:
$B = \dfrac{2008}{1}+ \dfrac{2007}{2} + \dfrac{2006}{3}+ ... + \dfrac{1}{2008} \\ = 2008 + (\dfrac{2008}{1}+ \dfrac{2007}{2}+ \dfrac{2006}{3}+ ... + \dfrac{1}{2008}) - 2008 \\ = {(1 + \dfrac{2008}{1}) + (1 + \dfrac{2007}{2}) + ... + (1 + \dfrac{1}{2008})} - 2008 \\ = (\dfrac{2009}{1} + \dfrac{2009}{2} + ... + \dfrac{2009}{2008}) - 2008 \\ = 2009(1 + \dfrac{1}{2} + ... + \dfrac{1}{2008}) - 2008 \\ = 2009(A +1 - \dfrac{1}{2009}) - 2008 \\ = 2009A + 2009 - 1 - 2008 \\ = 2009 \\ \rightarrow \dfrac{A}{B} = \dfrac{1}{2009}$

~> Chú ý đánh latex. Xem thêm tại Cách gõ công thức Toán, Vật lí, Hóa học
 
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