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)

\(A=\frac{9999}{32000}=0,31246875...\)

\(\frac{1}{1000}=0,001\Rightarrow0,31246875...>0,001\)

\(\Rightarrow A>\frac{1}{1000}\)

Gọi D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)

Số thừa số của C và D bằng nhau (đều bằng 100)

Ap dụng tính chất: a/b < 1 => a/b < a+m/b + m (b, m > 0)

Ta có:

\(\frac{1}{2}

????

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