K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

12 tháng 3 2018

Ta có : 

1/6 < 1/5 , 1/7 < 1/5 , ... 1/19 < 1/5

=> 1/6 + 1/7 + ...+ 1/19 < 1/5 + 1/5 + ...+ 1/5

=> 1/6 + 1/7 + ...+ 1/19  < 1/5 . 14 

=> 1/6 + 1/7 + ...+ 1/19 < 14/5 = 2 , 8 

1 tháng 7 2021

Đặt A = \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\) 

\(A=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+...+\frac{1}{19}\right)\) 

\(\Rightarrow A< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(\Rightarrow A< \frac{1}{5}\cdot5+\frac{1}{10}\cdot5+\frac{1}{15}\cdot5\)

\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow A< \frac{11}{6}< 2\) 

\(\Rightarrow A< 2\left(đpcm\right)\)

DD
21 tháng 5 2021

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

21 tháng 5 2021

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)

15 tháng 5 2023

Ta có : \(\dfrac{1}{6}>\dfrac{1}{10}\)

\(\dfrac{1}{7}>\dfrac{1}{10}\)

\(\dfrac{1}{8}>\dfrac{1}{10}\)

\(\dfrac{1}{9}>\dfrac{1}{10}\)

\(\dfrac{1}{10}=\dfrac{1}{10}\)

Cộng tất cả các vế ( phải theo phải ) ( trái theo trái ta được )

\(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{5}{10}=\dfrac{1}{2}\)

15 tháng 5 2023

Ta có:
\(\dfrac{1}{6}>\dfrac{1}{10}\)
\(\dfrac{1}{7}>\dfrac{1}{10}\)
\(\dfrac{1}{8}>\dfrac{1}{10}\)
\(\dfrac{1}{9}>\dfrac{1}{10}\)
\(\dfrac{1}{10}=\dfrac{1}{10}\)
Do đó ta có:
\(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}\times5\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{5}{10}=\dfrac{1}{2}\)
Vậy \(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{2}\)

11 tháng 4 2018

de lam ban oi!

11 tháng 4 2018

ta có

1/2<1/1.2

1/3<1/2.3

...

1/32<1/31.32

=>1/2+1/3+...+1/32<1/1.2+1/2.3+...+1/31.32

=>1/2+1/3+...+1/32<1/1-1/2+1/2-1/3+...+1/31-1/32

=>1/2+1/3+...+1/32<1/1-1/32=31/32

vì 31/32<1

=>tổng đó <1

ta lại có 1+1=2 mà 2 <3

=>tổng đó <3

vậy:-------(bn tự lm nha)

k cho mik vs nha

26 tháng 4 2016

Ta có:

        1/2^2 < 1/1.2

        1/3^2 < 1/2.3

         1/4^2< 1/3.4

     ........................

         1/8^2<1/7.8

 Vậy B < 1/1.2+1/2.3+1/3.4+....+1/7.8

B< 1-1/8

B<7.8<1

=> B<1     

12 tháng 4 2019

\(\frac{18}{18+19+20}>\frac{18}{18+19+20+21}\)

\(\frac{19}{18+19+21}>\frac{19}{18+19+20+21}\)

\(\frac{20}{18+19+21}>\frac{20}{18+19+20+21}\)

\(\frac{21}{18+19+21}>\frac{21}{18+19+20+21}\)

=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18}{18+19+20+21}+\frac{19}{18+19+20+21}+\frac{20}{18+19+20+21}+\frac{21}{18+19+20+21}\)

=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18+19+20+21}{18+19+20+21}\)

=>\(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>1\)

=>M>1

Còn lại mình không biết, đúng thì tick nhaoaoa