Help me pls!

Các dấu * là dấu nhân và các dấu / là dấu phân số.

ta có :
1/2 < 2/3
2/3 <3/4
.........
9999/10000 < 10000/10001
suy ra : A2 < 1/22/33/4*****9999/1000010000/10001
suy ra : A2 < 1/10001 < 1/10000= (1/100)2
suy ra A2 < (1/100)2 . Từ đó: A < 1/100

2 là mũ 2 nha bạn

Ta có: \(C=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\)

\(C=\dfrac{1\cdot3\cdot5...9999}{2\cdot4\cdot6...10000}\)

Gọi \(D=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)

Mà \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};\dfrac{5}{6}< \dfrac{6}{7};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

\(\Rightarrow\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)

\(\Rightarrow C< D\)

Ta lại có: \(C\cdot D=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\right)\)

\(\Rightarrow C\cdot D=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}...\dfrac{9999}{10000}\cdot\dfrac{10000}{10001}\)

\(\Rightarrow C\cdot D=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot6...9999\cdot10000}{2\cdot3\cdot4\cdot5\cdot6\cdot7...10000\cdot10001}\)

\(\Rightarrow C\cdot D=\dfrac{1}{10001}\)

Mà \(C< D\)

\(\Rightarrow C\cdot C< C\cdot D\)

\(\Rightarrow C\cdot C< \dfrac{1}{10001}\)

\(\Rightarrow C< \dfrac{1}{10001}\)

Mà \(\dfrac{1}{10001}< \dfrac{1}{100}\)

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

Vậy \(C< \dfrac{1}{100}\)

C = \(\dfrac{1}{2}\).\(\dfrac{3}{4}\).\(\dfrac{5}{6}\)....\(\dfrac{9999}{10000}\)

C < \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}....\dfrac{10000}{10001}\)

C² < \(\dfrac{1.\left(3.5.7...9999\right)}{\left(2.4.6...10000\right)}.\dfrac{\left(2.4.6...10000\right)}{\left(3.5.7...9999\right).10001}\)

C^{2 < \(\dfrac{1}{10001}\)}

C² < \(\left(\dfrac{1}{100}\right)^2\)

C < \(\dfrac{1}{100}\)

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

Chúc bạn học tốt !

ta có : 1/2 < 2/3

2/3 <3/4

......... 9999/10000 < 10000/10001 suy ra : A2 < 1/22/33/4*****9999/1000010000/10001 suy ra : A2 < 1/10001 < 1/10000= (1/100)2 suy ra A2 < (1/100)2 . Từ đó: A < 1/100 = 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

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 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)

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