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$\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)$

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HT.Phong (9A5) · Accepted Answer

$\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)$

$=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}$

$=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}$

$=\dfrac{10}{18}\cdot\dfrac{1}{4026042}$

$=\dfrac{5}{9}\cdot\dfrac{1}{4026042}$

$=\dfrac{5}{36234378}$Câu trả lời: $\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)$

$=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}$

$=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}$

$=\dfrac{10}{18}\cdot\dfrac{1}{4026042}$

$=\dfrac{5}{9}\cdot\dfrac{1}{4026042}$

$=\dfrac{5}{36234378}$