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:\(M=\frac{1}{2}\cdot\frac{3}{4}...\frac{9999}{10000}\) 

        \(N=\frac{2}{3}\cdot\frac{4}{5}...\frac{10000}{10001}\)

Dễ dàng nhận thấy: \(\frac{1}{2}

1/2.3/4.....9999/10000<1/100

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 < \(\frac{1}{100}\)

k nha bạn

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 \(C< \frac{1}{100}\left(đpcm\right)\)

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)

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) \(\Rightarrow A^2< \frac{1}{10001}\Rightarrow A< \sqrt{\left(\frac{1}{10001}\right)}< \sqrt{\left(\frac{1}{10000}\right)}=\frac{1}{100}\left(ĐPCM\right)\)

Ta có 27^5=3^3^5=3^15
243^3=3^5^3=3^15
Vậy A=B
2^300=2^(3.100)=2^3^100=8^100
3^200=3^(2.100)=3^2^100=9^100
Vậy A<B