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 A = (1/2)(3/4)(5/6) ... (9999/10000) (A > 0) 
.Và B = (2/3)(4/5)(6/7) ... (10000/10001) (B > 0) 
Ta có A.B = (1/2)(2/3)(3/4) ... (10000/10001) = 1/10001 (1) 
Mặt khác : 
1/2 < 2/3 
3/4 < 4/5 
................ 
................ 
9999/10000 < 10000/10001 
Nhân tất cả vế theo vế ---> A < B ---> A² < A.B (2) 
(1),(2) ---> A² < 1/10001 ---> A < căn(1/10001) < căn(1/10000) = 1/100 (đpcm)

nếu k^2=n thì ta nói căn bậc 2 của n là k(kEN)

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

M<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)

M²<\(\frac{1.\left(3.5.7....9999\right)}{\left(2.4.6....10000\right)}.\frac{\left(2.4.6....10000\right)}{\left(3.5.7....9999\right).10001}\)

Bạn rút gọn đi những phần mà mình đã đóng ngoặc nha

M²<\(\frac{1}{10001}\)

M²<\(\frac{1}{10000}\)

M²<\(\left(\frac{1}{100}\right)^2\)

=> M<1/100

jjjhhhhhhhhhhhhhh

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)

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

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

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

Dễ dàng nhận thấy: \(\frac{1}{2}

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

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

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)