\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)

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 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

Ta thấy : \(4=2^2;9=3^2;....;10000=100^2\) nên A có \(\left(100-2\right):1+1=99\) số hạng

Ta có :

\(\frac{3}{4}< \frac{4}{4}=1\)

\(\frac{8}{9}< \frac{9}{9}=1\)

\(\frac{15}{16}< \frac{16}{16}=1\)

\(......\)

\(\frac{9999}{10000}< \frac{10000}{10000}=1\)

\(\Rightarrow A=\frac{3}{4}+\frac{8}{9}+....+\frac{9999}{10000}< 1+1+...+1\)(Vì A có 99 số hạng nên cũng có 99 số 1 tương ứng)

\(\Rightarrow A< 99\)

\(A=\frac{3}{4}+\frac{8}{9}+...+\frac{9999}{10000}\)

\(A=1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{10000}\)

\(A=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)

Vì biểu thức trong dấu ngoặc đơn luôn lớn hơn 0 nên A<99

Vậy A<99

\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)

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

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

\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)

\(X=\frac{1}{100}.\frac{101}{2}\)

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

Study well 

Ta có :

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99\)\(\left(1\right)\)

gọi B là biểu thức trong ngoặc

Lại có :

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

\(\Rightarrow A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(1-\frac{1}{100}\right)>98\)

\(\Rightarrow A>98\)\(\left(2\right)\)

từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(98< A< 99\)

vậy A không phải là số tự nhiên

phần bạn đánh dấu (1) thì A<99 vì A= 99 trừ đi một số mà

\(\frac{3}{4}\)*\(\frac{8}{9}\)*\(\frac{15}{16}\)********\(\frac{9999}{10000}\)

= \(\frac{1\cdot3}{2^2}\)*\(\frac{2\cdot4}{3^2}\)********\(\frac{99\cdot101}{100^2}\)

= \(\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\)* \(\frac{3\cdot4\cdot5\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot100}\)

= \(\frac{1}{100}\)*\(\frac{101}{2}\)=\(\frac{101}{200}\)

Ta có: A = \(\frac{3}{8}\). \(\frac{8}{9}\).\(\frac{15}{16}\). ... .\(\frac{9999}{10000}\)
\(\Rightarrow\) A = \(\frac{1.3}{2^2}\).\(\frac{2.4}{3^2}\). \(\frac{3.5}{4^2}\). ... . \(\frac{99.101}{100^2}\)
\(\Rightarrow\) A = \(\frac{1.111}{2.100}\)= \(\frac{111}{200}\)
Vậy: A = \(\frac{111}{200}\).

\(A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\left(3\cdot5\right)\cdot.....\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right).......\left(100\cdot100\right)}\)

\(A=\frac{\left(1\cdot2\cdot3\cdot.....\cdot99\right)\left(3\cdot4\cdot5\cdot.....\cdot101\right)}{\left(2\cdot3\cdot4\cdot.....\cdot100\right)\left(2\cdot3\cdot4\cdot.....\cdot100\right)}\)

\(A=\frac{1\cdot101}{100\cdot2}\)

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

k mk nha mk nhanh nhất mk trả lời đúng 100%

\(M=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot....\cdot\frac{99\cdot101}{100\cdot100}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\frac{1\cdot101}{2\cdot100}=\frac{101}{200}\)Vậy M = \(\frac{101}{200}\)

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

\(M=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)