Câu hỏi của đạt Nguyễn tiến

Question

(1/2-1)(1/3-1)...(1/2008-1)(1/2009-1)

HT.Phong (9A5) · Accepted Answer

$\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2009}-1\right)\
=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)...\left(\dfrac{1}{2009}-\dfrac{2009}{2009}\right)\
=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\dfrac{-2008}{2009}\
=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2008}{2009}\
=\dfrac{2\cdot3\cdot...\cdot2008}{\left(2\cdot3\cdot...\cdot2008\right)\cdot2009}\
=\dfrac{1}{2009}$Câu trả lời: $\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2009}-1\right)\
=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)...\left(\dfrac{1}{2009}-\dfrac{2009}{2009}\right)\
=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\dfrac{-2008}{2009}\
=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2008}{2009}\
=\dfrac{2\cdot3\cdot...\cdot2008}{\left(2\cdot3\cdot...\cdot2008\right)\cdot2009}\
=\dfrac{1}{2009}$

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