Ta có: \(\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{9}\right)>\dfrac{1}{9}.6=\dfrac{6}{9}>\dfrac{1}{2}\)  (1)

\(\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)>\dfrac{1}{19}.10=\dfrac{10}{19}>\dfrac{1}{2}\)  (2)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>\left(1\right)+\left(2\right)\)

\(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{19}>1\left(đpcm\right)\)

1/4+1/5+1/6+...+1/19=(1/4+1/5+...+1/11)+(1/12+1/13+...+1/19)>(1/11+1/11+...+1/11)+(1/19+1/19+...+1/19)=8/11+8/19=240/209>209/209=1

⇒B>1

Tham khảo:

Ta có :

\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\)

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

Ta thấy :

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

\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}\)\(=\dfrac{1}{19}.5=\dfrac{5}{19}>\dfrac{1}{2}\)

⇒ \(B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}>1\)

⇒ \(B>1\) \(\left(đpcm\right)\)

đề là j vậy?

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

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 

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

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

\(\Rightarrow B>1\)

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

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

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