Giải:
Ta có:
\(\dfrac{5}{20}>\dfrac{5}{25}\) ; \(\dfrac{5}{21}>\dfrac{5}{25}\) ;\(\dfrac{5}{22}>\dfrac{5}{25}\) ; \(\dfrac{5}{23}>\dfrac{5}{25}\) ; \(\dfrac{5}{24}>\dfrac{5}{25}\)
\(\Rightarrow S=\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}>\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}=1\)
Vậy \(S=\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}>1\) ( đpcm )
Giải:
Dễ thấy:
\(20< 25\Leftrightarrow\dfrac{5}{20}>\dfrac{5}{25}\)
\(21< 25\Leftrightarrow\dfrac{5}{21}>\dfrac{5}{25}\)
\(.....................\)
\(24< 25\Leftrightarrow\dfrac{5}{24}>\dfrac{5}{25}\)
Cộng vế theo vế ta có:
\(S>\dfrac{5}{25}+\dfrac{5}{25}+...+\dfrac{5}{25}=\dfrac{5}{25}.5=\dfrac{25}{25}=1\)
Vậy \(S>1\) (Đpcm)
\(S=\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+\dfrac{5}{23}+\dfrac{5}{24}>\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}+\dfrac{5}{25}=1\)
\(\Rightarrow S>1\)