Sai đề. Sửa đề :v

Cmr: \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)

Giải:

Đặt \(A=\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\)

Ta có:

\(A=\dfrac{1}{5}+\left(\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}\right)+\left(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\right)\)

\(A< \dfrac{1}{5}\left(\dfrac{1}{14.3}\right)+\left(\dfrac{1}{61.3}\right)\)

\(A< \dfrac{1}{5}+\dfrac{3}{14}+\dfrac{3}{61}\)

\(A< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{1}{20}\)

\(A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow A< \dfrac{1}{2}\)

Vậy \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\) \((đpcm)\)

Ta có \(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{14}< \frac{1}{10};\frac{1}{28}< \frac{1}{10}\)

\(\frac{1}{44}< \frac{1}{40};\frac{1}{61}< \frac{1}{40};\frac{1}{85}< \frac{1}{40};\frac{1}{97}< \frac{1}{40}\)

\(\Rightarrow\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}=\frac{1}{5}+\frac{1}{5}+\frac{1}{10}=\frac{5}{10}=\frac{1}{2}\)\(\Rightarrow A< \frac{1}{2}\)