Lời giải:
Ta có:
\(\frac{1}{13}; \frac{1}{14}; \frac{1}{15}<\frac{1}{12}\)
\(\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}; \frac{1}{62};\frac{1}{63}< \frac{1}{60}\)
\(\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}\)
Do đó:
\(A< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{9}{20}+\frac{1}{20}\)
\(\Leftrightarrow A< \frac{1}{2}\) (đpcm)
Đặt biểu thức bằng A:
\(\Rightarrow A=\dfrac{1}{5}\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Ta thấy: \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{61}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{3}{31}+\dfrac{3}{61}< \dfrac{1}{2}\left(đpcm\right)\)