Đáp án:
$a)$
\[\bigg(1+\dfrac{1}{2}\bigg)\bigg(1+\dfrac{1}{5}\bigg)\bigg(1+\dfrac{1}{9}\bigg)...\bigg(1+\dfrac{1}{1224}\bigg)\] \[=\dfrac{3}{2}.\dfrac{6}{5}.\dfrac{10}{9}...\dfrac{1225}{1224}\] \[=\dfrac{6}{4}.\dfrac{12}{10}.\dfrac{20}{18}....\dfrac{2450}{2448}\] \[=\dfrac{2.3}{1.4}.\dfrac{3.4}{2.5}.\dfrac{4.5}{3.6}...\dfrac{49.50}{48.51}\] \[=\dfrac{2.3.3.4.4.5.5...49.49.50}{(2.3.4.5...48).(4.5...51)}=\dfrac{49}{17}\]
