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27 tháng 7 2021

\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)

\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(M=2.\dfrac{3}{16}\)

\(M=\dfrac{3}{8}\)

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)

3 tháng 4 2018

\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

\(\dfrac{3}{8}< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)

9 tháng 5 2018

bạn chép gì vậy????hay là não bạn có vấn đề?

23 tháng 6 2018

a, Ta có :

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

7 tháng 5 2017

lầy dạ??

4 tháng 5 2017

Ta có :

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét :

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

Giải

Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)

Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

D< 1 - \(\dfrac{1}{20}\)

D< \(\dfrac{19}{20}\)<1

\(\Rightarrow\)D< 1

Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1

30 tháng 4 2017

A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)

A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)

Ta có :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :

\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1

A<\(\dfrac{49}{200}< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\)

Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)

Ta lại có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)

Từ (1) và (2) suy ra đpcm.

1 tháng 4 2022

Hay quá

 

17 tháng 3 2017

b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=93-3=90\)

\(\Rightarrow x=90:2=45\)

19 tháng 3 2017

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