B = \(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)

    = \(\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{59.61}{60.60}\)

    = \(\frac{2.4.3.5.4.6...59.61}{3.3.4.4.5.5...60.60}\)

    = \(\frac{2.3....59}{3.4...60}.\frac{4.5...61}{3.4...60}\)

    = \(\frac{2}{60}.\frac{61}{3}\)= \(\frac{61}{90}\)

Chúc bạn học tốt!

\(B=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right)...\left(99.101\right)}{2^2.3^2.4^2.5^2...100^2}=\frac{\left(1.2.3.4...99\right).\left(3.4.5.6...101\right)}{\left(2.3.4.5...100\right)\left(2.3.4.5...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)

B = \(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}\frac{4.6}{5^2}...\frac{99.101}{100^2}=\frac{1.3.2.4.3.5.4.6...99.101}{2.2.3.3.4.4.5.5...100.100}\)

   =\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

Vật B = \(\frac{101}{200}\)

đúng cái đi   

\(B=\frac{8}{9}^.\frac{15}{16}^.\frac{24}{25}^........^.\frac{3599}{3600}\)

\(B=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}.....\frac{59.61}{60.60}\)

B = \(\left(\frac{2.3.4.....59}{3.4.5.....60}\right).\left(\frac{4.5.6.....61}{3.4.5.....60}\right)\)

\(B=\frac{2}{60}.\frac{61}{3}\)

B = \(\frac{61}{90}\)

B= \(\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{59.61}{60.60}\)

B=\(\frac{2.61}{3.60}\)

B= \(\frac{61}{90}\)

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

Ta có : \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

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

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{15}+...+\frac{1}{10000}\right)\)

\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99\)

\(\Rightarrow\)S<99 (1)

Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)

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

Ta có : \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

...

\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)

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

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

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

\(\Rightarrow\)S>99-1=98 (2)

Từ (1) và (2)

\(\Rightarrow\)98<S<99

\(\Rightarrow\)S\(\notin\)N

Vậy S\(\notin\)N.

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

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

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

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

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

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

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

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

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

Study well 

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

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

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

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

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%