Cho đoạn thẳng \(AB\). Gọi \(I\) là trung điểm của đoạn thẳng \(AB\) và điểm \(M\) là trung điểm của đoạn thẳng \(AI\). Biết rằng \(AM = 3\) cm. Khi đó, độ dài đoạn thẳng \(AB\) bằng

 Ta có \[A = \left( {1 + \frac{1}{{1.3}}} \right).\left( {1 + \frac{1}{{2.4}}} \right).\left( {1 + \frac{1}{{3.5}}} \right).\,\,...\,\,.\left( {1 + \frac{1}{{2022.2024}}} \right)\,\]

\[ = \frac{4}{{1\,\,.\,3}}.\frac{9}{{2\,.\,4}}.\,\frac{{16}}{{3\,.\,5}}.\,\,...\,\,.\frac{{4\,\,092\,\,529}}{{2022\,.\,2024}}\]

\[ = \frac{{2\,.\,2}}{{1\,\,.\,3}}.\frac{{3\,.\,3}}{{2\,.\,4}}.\frac{{4\,.\,4}}{{3\,.\,5}}.\,\,...\,\,.\frac{{2023\,.\,2023}}{{2022\,.\,2024}}\]

\[ = \frac{{2\,.\,3\,.\,4.\,\,...\,\,.2023}}{{1\,.\,2\,.\,3.\,\,...\,\,.2022}}.\frac{{2\,.\,3\,.\,4.\,\,...\,\,.2023}}{{3\,.\,4\,.\,5.\,\,...\,\,.2024}}\]

\[ = \frac{{2023}}{1}.\frac{2}{{2024}} = \frac{{2023}}{{1012}}\].

Vậy \(A = \frac{{2023}}{{1012}}\).

1.

a) \(\frac{5}{6} - \left( {\frac{1}{3} + \frac{1}{2}} \right):20\% \)\( = \frac{5}{6} - \left( {\frac{2}{6} + \frac{3}{6}} \right) \cdot 20\% \) 

\( = \frac{5}{6} - \frac{5}{6} \cdot \frac{1}{5}\)\( = \frac{5}{6} - \frac{1}{6}\)\( = \frac{4}{6} = \frac{2}{3}\);

b) \(\frac{3}{{17}}.\frac{6}{{29}} - \frac{3}{{17}}.\frac{{35}}{{29}} + 2022\frac{3}{{17}}\)\( = \frac{3}{{17}} \cdot \left( {\frac{6}{{29}} - \frac{{35}}{{29}}} \right) + 2022\frac{3}{{17}}\)

\( = \frac{3}{{17}} \cdot \left( {\frac{{ - 29}}{{29}}} \right) + 2022\frac{3}{{17}}\)\( = \frac{3}{{17}} \cdot \left( { - 1} \right) + 2022\frac{3}{{17}}\)

\( = \frac{{ - 3}}{{17}} + 2022\frac{3}{{17}} = 2022\).

2.