\(A=\frac{1}{2}\times\frac{3}{4}......\frac{9999}{10000}\)

Đặt : \(B=\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}.......\frac{10000}{10001}\)

Vì \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};.....\frac{9999}{10000}< \frac{10000}{10001}\)

Nên A<B  mà A>0; B>0

\(\Rightarrow A^2< A\times B=\left(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}.....\frac{9999}{10000}\right)\times\left(\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}......\frac{10000}{10001}\right)\)\(=\frac{1}{2}\times\frac{2}{3}\times\frac{4}{5}......\frac{9999}{10000}\times\frac{10000}{10001}\)\(=\frac{1}{10001}< \frac{1}{10000}=\frac{1}{100^2}=0.01^2\)\(\Rightarrow A^2< 0.01^2\)hay A < 0.01

A<2/3*4/5*6/7...10000/10001

A^2<A*(2/3*4/5*6/7...10000/10001)

A^2<\(\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6...9999\cdot10000}{2\cdot3\cdot4\cdot5\cdot6\cdot7...10000\cdot10001}\)

A^2<1/10001

0,01=1/100

1/100^2=1/10000

A^2<1/10001<1/10000

Đặt A = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{9998}{9999}.\frac{10000}{10000}\)

Rõ ràng A < A'

=> A² < A . A' \(=\frac{1}{10000}=\frac{1}{100^2}\)

Nên A < 0,01

Ta có \(\frac{2+3}{4+6}=\frac{1}{2}\)

\(\frac{2-3}{4-6}=\frac{-1}{-2}=\frac{1}{2}\)

=> \(\frac{1}{2}=\frac{1}{2}hay\frac{2+3}{4+6}=\frac{2-3}{4-6}\)

Mai lên lop tao giai cho 

Cách giải là:Đếu biết

< k nha

Đặt \(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{10000}{10001}\)

Ta có: \(S=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{10000}{10001}=B\)

Do đó: \(S\cdot S^2< S\cdot B\)

hay \(S^2< \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}=\frac{1}{10001}\)

⇔\(S^2< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{100}\cdot\frac{1}{100}=\left(0,01\right)^2\)

hay S<0,01(đpcm)

Ta có: \(0,01=\frac{1}{100}\)

Mà \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)

Ta thấy: \(\frac{1}{100}=\frac{100}{10000}\)

Vì \(\frac{9999}{10000}>\frac{100}{10000}hay\frac{9999}{10000}>\frac{1}{100}\)

Nên \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}>\frac{1}{100}hay\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}>0,01\)

Vậy \(A>0,01\)

Ta có: \(0,01=\frac{1}{100}\)

Đặt  \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)

Xét  \(AB=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)

\(\Leftrightarrow\)\(AB=\frac{1.2.3.4.5.6.....9999.10000}{2.3.4.5.6.7.....10000.10001}\)

\(\Leftrightarrow\)\(AB=\frac{1}{10001}\)

​Vì A < B

\(\Rightarrow\)A² < AB

\(\Rightarrow A^2< \frac{1}{10001}< \frac{1}{10000}\)

\(\Rightarrow A< \frac{1}{100}hayA< 0,01\)

Vậy A < 0,01