Ta có : \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};....;\frac{99}{100}< \frac{100}{101}\)

Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)\(\Rightarrow B>A\)

\(\Rightarrow A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(\Rightarrow A.B=\frac{1}{101}\)

Vì \(B>A\)\(\Rightarrow A.B>A.A=A^2\)

\(\Rightarrow\frac{1}{101}>A^2\)

Mà \(\frac{1}{10^2}>\frac{1}{101}>A^2\Rightarrow\frac{1}{10^2}>A^2\)

\(\Rightarrow\frac{1}{10}< A\left(1\right)\)\(\)

Ta lai có :

\(\frac{1}{2}=\frac{1}{2};\frac{3}{4}>\frac{2}{3};\frac{5}{6}>\frac{4}{5};...;\frac{99}{100}>\frac{98}{99}\)

Đặt \(C=\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A.C=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)

\(\Rightarrow A.C=\frac{1}{2}.\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)

\(\Rightarrow A.C=\frac{1}{200}\)

Vì \(A>C\)

\(\Rightarrow A^2>A.C=\frac{1}{200}\)

Mà \(A^2>\frac{1}{200}>\frac{1}{15^2}\)

\(\Rightarrow A^2>\frac{1}{15^2}\)

\(\Rightarrow A>\frac{1}{15}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{1}{15}< A< \frac{1}{10}\)

\(\RightarrowĐPCM\)

                                                                    Bài giải

 \(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)

\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)

\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)

\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)

\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)

\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)

\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)

\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)

\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)

\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)

\(A\cdot C=\frac{1}{200}\)

\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)

\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)

\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)

\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)

\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)

\(\Rightarrow\text{ }\text{ĐPCM}\)

                                                                    Bài giải

 \(\frac{1}{2}< \frac{2}{3}\text{ ; }\frac{3}{4}< \frac{4}{5}\text{ ; }\frac{5}{6}< \frac{6}{7}\text{ ; }...\text{ ; }\frac{99}{100}< \frac{100}{101}\)

\(\text{Đặt }B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)

\(\Rightarrow\text{ }A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)

\(\Rightarrow\text{ }A\cdot A< A\cdot B=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right)\)

\(A\cdot A< A\cdot B=\frac{1}{101}< \frac{1}{10}\)

\(A^2< \frac{1}{10}\text{ }\Rightarrow\text{ }A< \frac{1}{10}^{^{\left(1\right)}}\)

\(\frac{1}{2}=\frac{1}{2}\text{ ; }\frac{3}{4}>\frac{2}{3}\text{ ; }\frac{5}{6}>\frac{4}{5}\text{ ; }...\text{ ; }\frac{99}{100}>\frac{98}{99}\)

\(\text{Đặt }C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)

\(A\cdot C=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right)\cdot\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\right)\)

\(A\cdot C=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)

\(A\cdot C=\frac{1}{200}\)

\(\text{Vì }A>C\text{ }\Rightarrow\text{ }A^2>A\cdot C=\frac{1}{200}\)

\(\text{Mà }A^2>\frac{1}{200}>\frac{1}{15^2}\)

\(\Rightarrow\text{ }A>\frac{1}{15}^{^{\left(2\right)}}\)

\(\text{Từ }^{\left(1\right)}\text{ và }^{\left(2\right)}\)

\(\Rightarrow\text{ }\frac{1}{15}< A< \frac{1}{10}\)

\(\Rightarrow\text{ }\text{ĐPCM}\)

mình nghĩ kí hiệu đó là nhân 

Nếu là nhân thì lấy máy tính ra mà tính

Chúc bn hok tốt !

Kí hiệu * là gì vậy bn 

A= 1/2 >1/10 

1/2<1/10

1/2 < 1/10

Bạn ơi hình như là sai đề bạn ạ

sai ở đâu vậy

nhanh lên các bạn ơii