chứng minh
1/20,13 + 1 / 23,26 + 1 / 26,29 + ..... + 1 / 77,80 1/9
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\(\left(20,13-20,13:11+\dfrac{1}{4}\right):\dfrac{1}{4}\\ =\left(20,13-1,83+\dfrac{1}{4}\right):\dfrac{1}{4}\\ =\left(18,3+0,25\right):0,25\\ =18,55:0,25\\ =74,2\)
(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 :
\(\frac{1}{5}< \frac{6}{15}< \frac{12}{20}< \frac{12}{12}< \frac{13}{6}\)
`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:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)
\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)
\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)
\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)
\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)