\(S^2=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\\ \text{Ta có:}\\ \dfrac{1}{2}< \dfrac{2}{3}\\ \dfrac{3}{4}< \dfrac{4}{5}\\ \dfrac{5}{6}< \dfrac{6}{7}\\ ...\\ \dfrac{199}{200}< \dfrac{200}{201}\\ \Rightarrow S^2< \left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{200}{201}\right)\\ \Leftrightarrow S^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{200}{201}\\ \Leftrightarrow S^2< \dfrac{1\cdot2\cdot3\cdot...\cdot200}{2\cdot3\cdot4\cdot...\cdot201}\\ \Leftrightarrow S^2< \dfrac{1}{201}< \dfrac{1}{200}\)

Vậy ...

a, tổng các tử và mẫu mỗi phân sô trên đều bằng 200

b, \(A=\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

\(A=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)

\(A=200\left(\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2}+\dfrac{1}{200}\right)\)(đpcm)

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)

Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)

Ta có :

\(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{199}{200}< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{199}{200}.\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1.2.3.4....199.200}{2.3.4.5....200.201}=\dfrac{1}{201}\)

\(\Rightarrow\left(đpcm\right)\)

a) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=\dfrac{40}{14}=\dfrac{20}{7}\)

b) \(3\dfrac{5}{6}+2\dfrac{1}{6}x6=\dfrac{23}{6}+\dfrac{13}{6}x6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)

Ta có :

\(\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

\(=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)\left(\dfrac{199}{1}+1\right)-199\)\(=\dfrac{200}{199}+\dfrac{200}{199}+...+\dfrac{200}{2}+200-199\)

\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)

\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)\)

\(=200.A\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{200}\)

\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)

\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)

\(=-\dfrac{37}{16}\)

\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)

\(=\dfrac{5}{17}+\dfrac{-3}{17}\)

\(=\dfrac{2}{17}\)

\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)

\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)

\(=2-\dfrac{7}{30}\)

\(=\dfrac{53}{30}\)

\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)

\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)

\(=\dfrac{15}{34}\)