Giải:

\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\) 

\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\) 

\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\) 

\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\) 

\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\) 

Ta có:S=1/50+1/51+1/52+...+1/99

S>1/50+1/50+1/50+....+1/50(50 số hạng)

S>1/50x50

S>1>1/2

=>S>1/2

S = 1 / 50 + 1 / 51 +...+ 1 / 99 > 1 / 99 + 1 / 99 +...+ 1 / 99 = 50 / 99 > 50 / 100 = 1/2

\(S=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(có 50 số hạng)\(=\frac{50}{100}=\frac{1}{2}\)

Vậy \(S>\frac{1}{2}\) .

Có: \(\frac{1}{50}>\frac{1}{100}\\ \frac{1}{51}>\frac{1}{100}\\ \frac{1}{52}>\frac{1}{100}\\ .\\ .\\ .\\ \frac{1}{98}>\frac{1}{100}\\ \frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)(có 50 số hạng \(\frac{1}{100}\))

\(\Rightarrow S>\frac{1}{100}\cdot50\)

\(\Rightarrow S>\frac{50}{100}\)

\(\Rightarrow S>\frac{1}{2}\left(đpcm\right)\)

\(S=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\) (có 50 p/s \(\frac{1}{100}\))

\(\Rightarrow S>\frac{50}{100}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}\)

\(\Rightarrow\) đpcm

\(S=\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{99}>\frac{1}{99}+\frac{1}{99}+...+\frac{1}{99}=\frac{50}{99}>\frac{50}{100}=\frac{1}{2}\)

Ta có:

\(\frac{1}{2}=\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 PS)

Vì \(\frac{1}{50}>\frac{1}{100}\)

     \(\frac{1}{51}>\frac{1}{100}\)

      \(\frac{1}{52}>\frac{1}{100}\)

       .......................

      \(\frac{1}{99}>\frac{1}{100}\)

\(=>\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\right)>\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)                                                                                                    ( có 50 PS)

\(=>\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}>\frac{1}{2}\)

Tổng S có : (99 - 50) : 1 + 1 = 50 (số)\(S=\frac{1}{50}+\frac{1}{51}+....+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}+\frac{1}{100}\)(50 phân số \(\frac{1}{5}\)) = \(\frac{1}{100}.50=\frac{1}{2}\)

Vậy S > \(\frac{1}{2}\)

S = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)

Kết luận vậy S > 1/2

S=1/2

S>\(\frac{1}{2}\)

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ngu vảy 

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)

\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Ta có đpcm