Giải:
\(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}\)
Đặt \(B=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{10000}{10001}\)
Do \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)
Nên \(C< B\) Mà \(\left\{{}\begin{matrix}C>0\\B>0\end{matrix}\right.\)
\(\Rightarrow C^2< C.B=\left(\dfrac{1}{2}.\dfrac{3}{4}...\dfrac{9999}{10000}\right)\)\(\left(\dfrac{2}{3}.\dfrac{4}{5}...\dfrac{10000}{10001}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{9999}{10000}.\dfrac{10000}{10001}\)
\(=\dfrac{1.2.3.4.5.6...9999.10000}{2.3.4.5.6.7....10000.10001}\)
\(=\dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(\dfrac{1}{100}\right)^2\)
\(\Rightarrow C^2< \left(\dfrac{1}{100}\right)^2\Leftrightarrow C< \dfrac{1}{100}\)
Vậy \(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\) (Đpcm)
