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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
Đặt \(I=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
Ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};.....;\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< D\)
Lại có: \(C\cdot D=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\right)\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{10001}\)
Mà C<D \(\Rightarrow C\cdot C< C\cdot D\)
Hay \(C\cdot C< \frac{1}{10001}\)
\(\Rightarrow C< \frac{1}{10001}< \frac{1}{100}\)
Vậy \(C< \frac{1}{100}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt:\(M=\frac{1}{2}\cdot\frac{3}{4}...\frac{9999}{10000}\)
\(N=\frac{2}{3}\cdot\frac{4}{5}...\frac{10000}{10001}\)
Dễ dàng nhận thấy: \(\frac{1}{2}
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt :\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
\(N=\frac{2}{3}.\frac{4}{5}...\frac{10000}{10001}\)
Ta thấy:\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};....;\frac{9999}{10000}< \frac{10000}{10001}\)
Mặt khác ta thấy:
\(C.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)
\(C.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{9999}{10000}.\frac{10000}{10001}\)
\(C.N=\frac{1.2.3....9999.10000}{2.3.4....10000.10001}\)
Rút gọn phép tính \(C.N\)
\(C.N=\frac{1}{10001}\)
\(C.C< N\Rightarrow C.C< C.N\)
Hay\(C.C< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{10}.\frac{1}{10}\)
\(\Rightarrow C< \frac{1}{10000}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt :
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)
Đặt :
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{9998}{9999}.\frac{10000}{10000}\)
Ta thấy " A<B
\(\Rightarrow A.A< A.B=\frac{1}{100^2}\\ \Rightarrow A^2< \frac{1}{100^2}\\ \Rightarrow A< \frac{1}{100}\)
Đặt \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)\(\left(A>0\right)\)
.Và \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\)\(\left(B>0\right)\)
Mặt khác :
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
... ... ...
\(\frac{9999}{10000}< \frac{10000}{10001}\)
Nhân tất cả vế theo vế \(\Rightarrow A< B\Rightarrow A^2< A.B\left(2\right)\)
(1),(2) \(\Rightarrow A^2< \frac{1}{10001}\Rightarrow A< \sqrt{\left(\frac{1}{10001}\right)}< \sqrt{\left(\frac{1}{10000}\right)}=\frac{1}{100}\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.............\frac{10000}{10001}=M\)
=> A.A < A.M = \(\frac{1}{10001}\)
=> A2 < \(\frac{1}{10000}=\left(\frac{1}{100}\right)^2\)
=> A < \(\frac{1}{100}\)
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