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9 tháng 5
Ta có thể chia tổng \(B\) thành các nhóm như sau: \(B = \left(\right. \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \left.\right) + \left(\right. \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} \left.\right) + \left(\right. \frac{1}{16} + \frac{1}{17} + \frac{1}{18} + \frac{1}{19} \left.\right)\) Ta sẽ đánh giá từng nhóm:
  1. Nhóm 1: \(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} > \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{4}{7} > \frac{4}{8} = \frac{1}{2}\)
  2. Nhóm 2: \(\frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} > \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} = \frac{8}{15} > \frac{8}{16} = \frac{1}{2}\)
  3. Nhóm 3: \(\frac{1}{16} + \frac{1}{17} + \frac{1}{18} + \frac{1}{19} > \frac{1}{19} + \frac{1}{19} + \frac{1}{19} + \frac{1}{19} = \frac{4}{19}\)
Vậy, \(B > \frac{1}{2} + \frac{1}{2} + \frac{4}{19} = 1 + \frac{4}{19} > 1\) Vậy, \(B > 1\) (điều phải chứng minh). Kết luận: Ta đã chứng minh được rằng \(B > 1\).
9 tháng 5

B = 1/4 + 1/5 + 1/6 + ... + 1/19

= (1/4 + 1/5) + (1/6 + 1/7) + (1/8 + 1/9) + ... + (1/18 + 1/19)

= 0.25 + 0.2 + 0.167 + ... + 0.053

= 1.767

17 tháng 4 2024

7 tháng 5 2016

Xét: 1 / 4 > 1 / 16 ; 1 / 5 > 1 / 16 

=) 1 / 4 + 1 / 5 + 1 / 6 + ... + 1 / 19  > 16 . 1 / 16 = 16 /16 = 1

=) B > 1

Vậy B > 1

B= 1/4+(1/5+1/6+...+1/9)+(1/10+1/11+...+1/19)
Vì 1/5+1/6+...+1/9 > 1/9+1/9+...+1/9 nên 1/5+1/6+...+1/9 > 5/9 >1/2
Vì 1/10+1/11+...+1/19 > 1/19+1/19+...+1/19 nên 1/10+1/11+...+1/19 > 10/19 >1/2
Suy ra: B > 1/4+1/2+1/2 > 1

17 tháng 4 2024

21 tháng 4 2016

B = 1/4 + 1/5 + 1/6 +....+1/19

> 1/4 + ( 1/20 + 1/20 +.....+1/20) ( 15 p/s 1/20) = 1/4 + 3/4 = 1

=> B > 1

Vậy B > 1 

29 tháng 4 2017

Ta có :

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)

\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)

Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)

Vì \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)

\(\Rightarrow B>1\)

29 tháng 4 2017

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)

\(\Rightarrow B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\right)\)Do \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)

\(\Rightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>4.\frac{1}{8}=\frac{1}{2}\left(1\right)\)

\(\Rightarrow\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}>4.\frac{1}{12}=\frac{1}{3}\left(2\right)\)

\(\Rightarrow\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}>4.\frac{1}{16}=\frac{1}{4}\left(3\right)\)

\(\Rightarrow\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}>4.\frac{1}{20}=\frac{1}{5}\left(4\right)\)

Từ (1) , ( 2 ) , ( 3 ) và ( 4 ) suy ra :

\(B>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\) 

\(B>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{5}\)

\(B>\frac{1}{2}+\frac{1}{2}+\frac{1}{5}\)

\(B>1+\frac{1}{5}\Rightarrow B>1\)

Vậy : \(B>1\)

23 tháng 5 2019

#)Giải :

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(B=1-\frac{1}{5}< 1\)

\(\Leftrightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(\Leftrightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}< 1\)

\(\Leftrightarrow B< 1\)

         #~Will~be~Pens~#

23 tháng 5 2019

Ta có : \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)

\(B=\frac{1}{4}+\left[\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right]+\left[\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right]\)

Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}=\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)

\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}=\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}\)

\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{2}{4}+\frac{2}{4}\)

\(\Rightarrow B>\frac{5}{4}>\frac{4}{4}=1\)

Vậy B > 1

17 tháng 4 2024

17 tháng 4 2024

3 tháng 4 2018

Ta có : 

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\)

\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\)

\(\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\right)\)

\(B>\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\left(\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\right)+\)

\(\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\right)\)

\(B>\frac{4}{8}+\frac{4}{12}+\frac{4}{16}+\frac{4}{20}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{77}{60}>\frac{60}{60}=1\)

\(\Rightarrow\)\(B>1\)

Vậy \(B>1\)

Chúc bạn học tốt ~