Giải:
Ta có:
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)\) \(+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)\)
Nhận xét:
\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\) \(=\dfrac{5}{9}>\dfrac{5}{10}=\dfrac{1}{2}\)
\(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+...+\dfrac{1}{19}\) \(=\dfrac{10}{19}>\dfrac{1}{2}\)
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=1+\dfrac{1}{4}>1\)
Vậy \(B>1\) (Đpcm)