C=1/2x3/4x5/6x......x9999/10000<1/100
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\(C=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{9999}{10000}\)(1)
Ta có : \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< \frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}\times...\times\frac{10000}{10001}\)(2)
Từ (1) và (2) \(\Rightarrow C^2< \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times...\times\frac{9999}{10000}\times\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}=\left(\frac{1}{100}\right)^2\)
\(\Rightarrow C< \frac{1}{100}\)(đpcm)
\(A=\frac{9999}{32000}=0,31246875...\)
\(\frac{1}{1000}=0,001\Rightarrow0,31246875...>0,001\)
\(\Rightarrow A>\frac{1}{1000}\)
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)