\[\begin{array}{l}
A = \left( {1 - \frac{1}{{1 + 2}}} \right)\left( {1 - \frac{1}{{1 + 2 + 3}}} \right)\left( {1 - \frac{1}{{1 + 2 + 3 + 4}}} \right).....\left( {1 - \frac{1}{{1 + 2 + 3 + ... + 2006}}} \right)\\
= \left( {1 - \frac{1}{{\frac{{2.3}}{2}}}} \right)\left( {1 - \frac{1}{{\frac{{3.4}}{2}}}} \right)\left( {1 - \frac{1}{{\frac{{4.5}}{2}}}} \right).....\left( {1 - \frac{1}{{\frac{{2006.2007}}{2}}}} \right)\\
= \left( {1 - \frac{2}{{2.3}}} \right)\left( {1 - \frac{2}{{3.4}}} \right)\left( {1 - \frac{2}{{4.5}}} \right).....\left( {1 - \frac{2}{{2006.2007}}} \right)\\
= \left( {\frac{{2.3 - 2}}{{2.3}}} \right)\left( {\frac{{3.4 - 2}}{{3.4}}} \right)\left( {\frac{{4.5 - 2}}{{4.5}}} \right).....\left( {\frac{{2006.2007 - 2}}{{2006.2007}}} \right)\\
n(n + 1) - 2 = n(n + 2 - 1) + n - (n + 2) = n(n + 2 - 1 + 1) - (n + 2) = n(n + 2) - (n + 2) = (n + 2)(n - 1)\\
Nen\,\,A = \frac{{1.4}}{{2.3}}.\frac{{2.5}}{{3.4}}.\frac{{3.6}}{{4.5}}.....\frac{{2005.2008}}{{2006.2007}} = \frac{{(1.2.3.....2005)(4.5.6.....2008)}}{{(2.3.4.....2006)(3.4.5....2007)}} = \frac{{2008}}{{2006.3}} = \frac{{1004}}{{3009}}\\
Vay\,\,A = \frac{{1004}}{{3009}}
\end{array}\]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn