\(\left(20,13-20,13:11+\dfrac{1}{4}\right):\dfrac{1}{4}\\ =\left(20,13-1,83+\dfrac{1}{4}\right):\dfrac{1}{4}\\ =\left(18,3+0,25\right):0,25\\ =18,55:0,25\\ =74,2\)

(20,13−20,13:11+14):14=(20,13−1,83+14):14=(18,3+0,25):0,25=18,55:0,25=74,2

Ta có:A=20,13.100+2013.100/50+201,3:0,1+2,013:0,001

        A=2013+2013.2+2013+2013

        A=2013.(1+2+1+1)

        A=2013.5

        A=10065

Ta có:A=20,13.100+2013.100/50+201,3:0,1+2,013:0,001

           A=2013+2013.2+2013+2013

           A=2013.(1+2+1+1)

           A=2013.5

           A=10065

a, Ta có :

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)

thứ tự từ bé đến lớn là : 

\(\frac{1}{5}< \frac{6}{15}< \frac{12}{20}< \frac{12}{12}< \frac{13}{6}\)

Ta có: \(\frac14=\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac19=\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{100}<\frac{1}{10^2}<\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac14+\frac19+\cdots+\frac{1}{100}<1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\) (1)

Ta có: \(\frac14=\frac{1}{2^2}>\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac19=\frac{1}{3^2}>\frac{1}{3\cdot4}=\frac13-\frac14\)

....

\(\frac{1}{100}=\frac{1}{10^2}>\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)

Do đó: \(\frac14+\frac19+\cdots+\frac{1}{100}>\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{10}-\frac{1}{11}=\frac12-\frac{1}{11}=\frac{9}{22}\) (2)

Từ (1),(2) suy ra \(\frac{9}{22}<\frac14+\frac19+\cdots+\frac{1}{100}<\frac{9}{10}\)

`Answer:`

 \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)

a) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)

\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)

b) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)

\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)

\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)

\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)

\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)