Tính để mai làm mà thoii kệ, làm luôn :vv
Giải:
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+..+\dfrac{1}{19}\)
\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)\)
Vì \(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\)
Nên \(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}>\dfrac{5}{9}>\dfrac{1}{2}\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+...+\dfrac{1}{19}\)
Nên \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{10}{19}>\dfrac{1}{2}\)
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(\Rightarrow B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}>1\)
Giải:
Ta có: \(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}\right)\)
Dễ thấy:
\(5< 20\Leftrightarrow\dfrac{1}{5}>\dfrac{1}{20}\)
\(6< 20\Leftrightarrow\dfrac{1}{6}>\dfrac{1}{20}\)
\(....................\)
\(19< 20\Leftrightarrow\dfrac{1}{19}>\dfrac{1}{20}\)
Cộng vế theo vế ta có:
\(B>\dfrac{1}{4}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có \(15\) phân số \(\dfrac{1}{20}\))
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{20}.15=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Vậy \(B>1\) (Đpcm)
