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

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

Lấy A.B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}=\frac{1}{10001}\)

mặt khác

Ta có

\(\frac{1}{2}< \frac{2}{3}\\\)

\(\frac{3}{4}< \frac{4}{5}\)

  ....

\(\frac{9999}{10000}< \frac{10000}{10001}\)

=> A<B

=> A.A<A.B

=>A²<\(\frac{1}{10001}< \frac{1}{10000}\)

=>A<\(\sqrt{\frac{1}{10000}}=\frac{1}{100}\)

Vậy \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)<\(\frac{1}{100}\)

ĐPCM

cái dấu\(\sqrt{ }\) mik chưa học bạn sửa cái chỗ gần về sau hộ mik nhé

đó là dấu căn bậc 2 bạn nhé :)) 

VD\(\sqrt{9}=3\\\) (3²=9)

\(\sqrt{16}=4\left(4^2=16\right)\)

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

Đặt C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)\(\left(C>0\right)\)

Và D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)\(\left(D>0\right)\)

Ta có :

C .D = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}\)\(=\frac{1}{10001}\)\(\left(1\right)\)

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 - - - > C < D - - - > C² < C . D \(\left(2\right)\)

\(\left(1\right),\left(2\right)\)- - - >C² < \(\frac{1}{10001}\)- - - >  C < căn \(\frac{1}{10001}\)< căn \(\frac{1}{10000}\)= \(\frac{1}{100}\)( đpcm )

lon hon la cai chac

A lon hon B

So sánh:

Ta có:

A > 0,01

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

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