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Ta có\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\)<\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\)=\(\frac{5}{6}\)(6 c/s \(\frac{1}{5}\))
Ta lại có \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{17}\)<\(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\)=\(\frac{7}{11}\)(7 c/s \(\frac{1}{11}\))
Suy ra \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)<\(\frac{110}{55}\)=2
Vậy...
Hok tốt
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< \frac{6}{5}+\frac{7}{11}\)
\(\Rightarrow A< \frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\)( ĐPCM )
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+....+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\left(1\right)\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\left(2\right)\)
Từ (1)(2) \(\Rightarrow A< \frac{6}{5}+\frac{7}{11}=\frac{66}{55}+\frac{35}{55}=\frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2\left(đpcm\right)\)
Ta có:
1/5=1/5
1/6<1/5
1/7<1/5
..........
1/10<1/5
=>1/5+1/6+...+1/10<1/5.6=6/5(1)
Lại có :
1/11=1/11
1/12<1/11
1/13<1/11
.............
1/17<1/11
=>1/11+1/12+1/13+...+1/17<1/11.7=7/11(2)
Từ (1)và (2)=>1/5+1/6+...+1/17<6/5+7/11=101/55<110/55=2
=>1/5+1/6+...+1/17<2
ĐPCM
◥ὦɧ◤ᗰIᑎᕼ™ᐯY™=ε/̵͇̿̿/'̿'̿ ̿ ̿̿ ̿̿ ̿̿
theo tớ nghĩ:
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}.200=\frac{2}{3}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}\)
Ta có:
\(\frac{1}{101}>\frac{1}{300}\)
\(\frac{1}{102}>\frac{1}{300}\)
..........................
\(\frac{1}{299}>\frac{1}{300}\)
\(\frac{1}{300}=\frac{1}{300}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)
\(\Rightarrow VT>200.\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\) (ĐPCM)
ta có
\(\frac{1}{300}< \frac{1}{101}\); \(\frac{1}{300}< \frac{1}{102}\); \(\frac{1}{300}< \frac{1}{102}\)....\(\frac{1}{300}< \frac{1}{299}\)
\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}< \frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)
\(\frac{200}{300}< \frac{1}{101}+\frac{1}{102}+...+\text{}\text{}\)
rút gọn là xong
ta có \(A=\frac{1}{100}+\frac{1}{101}+...+\frac{1}{149}\)
ta thấy \(\frac{1}{100}=\frac{1}{100}\)
\(\frac{1}{101}<\frac{1}{100}\)
\(\frac{1}{102}<\frac{1}{100}\)
................................
\(\frac{1}{149}<\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{149}<\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
\(=\frac{49}{100}<\frac{1}{2}\)
vì \(A<\frac{49}{100}<\frac{1}{2}\Leftrightarrow A<\frac{1}{2}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)( có 200 số )
Ta có
\(\frac{1}{101}>\frac{1}{300}\); \(\frac{1}{102}>\frac{1}{300}\); ...;\(\frac{1}{299}>\frac{1}{300}\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}+\frac{1}{300}\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}.200\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{2}{3}\)( dpcm )
Ta có\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200.\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\Rightarrowđpcm\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
có \(\frac{1}{2\cdot3}< \frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot4}< \frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot5}< \frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{9\cdot10}< \frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}>A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{8}{9}>A>\frac{2}{5}\)
Bạn ơi, sai rồi, mình k nhầm
làm sao mà \(\frac{1}{2^2}< \frac{1}{1.2}\)được
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>200.\frac{1}{300}\)
\(>\frac{2}{3}\)
Ta có
\(\frac{1}{101}>\frac{2}{3}\)
\(\frac{1}{102}>\frac{2}{3}\)
.
.
.
\(\frac{1}{300}>\frac{2}{3}\)
Vậy \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
Ta đặt \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}\)
Vì \(\frac{1}{101}>\frac{1}{102}>...>\frac{1}{299}>\frac{1}{300}\)
\(\Rightarrow A=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)
\(\Rightarrow A>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)
\(\Rightarrow A>\left(\frac{1}{200}\cdot100\right)+\left(\frac{1}{300}\cdot100\right)\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A>\frac{5}{6}>\frac{2}{3}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}\)>\(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)=\(\frac{1}{2}\)(có 200 c/s\(\frac{1}{200}\))
\(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\)>\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)=\(\frac{2}{3}\)(có 200 c/s \(\frac{1}{300}\))
Vậy \(\frac{1}{101}+\frac{1}{102+}+....+\frac{1}{300}\)>\(\frac{1}{2}+\frac{2}{3}=\frac{2}{3}\) Đpcm
Hok tốt