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 :
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}+\dfrac{1}{9}\)\(=\dfrac{1}{9}.5=\dfrac{5}{9}>\dfrac{1}{2}\)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}+\dfrac{1}{19}\)\(=\dfrac{1}{19}.5=\dfrac{5}{19}>\dfrac{1}{2}\)
⇒ \(B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}>1\)
⇒ \(B>1\) \(\left(đpcm\right)\)
