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

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

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

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

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%

\(8\frac{4}{17}-\left(2\frac{5}{9}+3\frac{4}{17}\right)=\frac{140}{17}-\left(\frac{23}{9}+\frac{55}{17}\right)=\frac{140}{17}-\frac{886}{153}=\frac{22}{9}=2,444444444444\)

\(\frac{3}{2^2}.\frac{2^3}{3^2}.\frac{5.3}{4^2}.......\frac{3.3333}{100^2}\)

mk ko nghĩ ra phần sau chắc là rút gọn thì phải!!?

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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 

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

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

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

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