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

Ta có:

\(\frac{1}{51}>\frac{1}{100}\)

\(\frac{1}{52}>\frac{1}{100}\)

...

\(\frac{1}{99}>\frac{1}{100}\)

\(\frac{1}{100}=\frac{1}{100}\)

=> S = \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)

Mà số số hạng của S là: (100 - 51) : 1 + 1 = 50 (số)

=> S \(>\frac{1}{100}.50\)

=> S \(>\frac{1}{2}\)

Vậy S > 1/2.

Ta có : 

\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=50.\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

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

Vậy \(S>\frac{1}{2}\)

Chúc bạn học tốt ~

\(S>\frac{1}{100}\cdot50=\frac{1}{2}\)

vì\(\frac{1}{51}>\frac{1}{52}>....>\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)> 50x1/100=1/2(dpcm)

ta có 1/51>1/100

        1/52>1/100

        ..................

        1/100=1/100

\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2

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

cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá

chúc bạn học tốt~

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

\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{2}\)

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

\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{50}{100}=\frac{1}{2}\left(đpcm\right)\)

\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{2}\)

\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 cái như z)

\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{50}{100}=\frac{1}{2}\left(đpcm\right)\)

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

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