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

Mai lên lop tao giai cho 

Cách giải là:Đếu biết

Câu 1:

A = (4/9 + 5/6) : (7/8 - 7/12)

A = (8/18 + 15/18) :(21/24 - 14/24)

A = 23/18 : 7/24

A = 23/18 x 24/7

A = 92/21

Câu B = 3/4.8/9.15/16...9999/10000

B = \(\frac{1.3}{2.2}\).\(\frac{2.4}{3.3}\)....\(\frac{99.101}{100.100}\)

B = \(\frac{1.2.3\ldots99}{2.3\ldots100}\). \(\frac{3.4.\ldots101}{2.3\ldots100}\)

B = \(\frac{1}{100}\) x \(\frac{101}{2}\)

B = \(\frac{101}{200}\)

a) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\)

    Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)

               \(\frac{1}{3^2}< \frac{1}{2.3}\)

                .................

             \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)

\(\Rightarrow B< 1-\frac{1}{2014}< 1\)

\(\Rightarrow B< 1\)

\(\Rightarrow1+B< 1+1\)

Hay \(A< 2\)

C) Ta có: \(\frac{1}{2}< \frac{2}{3}\)

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

                .................

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

\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\)

\(\Rightarrow C^2< \left(\frac{1}{2}.\frac{3}{4}.....\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\right)\)

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

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

\(\Rightarrow C< \frac{1}{100}\)

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

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