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 1/50>1/100    

         1/51>1/100

       ..........

          1/99>1/100

  vậy S>1/100*50=1/2

suy ra S>1/2

Tổng S có 50 phân số

=> S > 1/100 + 1/100 + 1/100 +...+ 1/100 (50 phân số) => S > 1/2.

Vậy S > 1/2

Tổng S có 50 phân số

=> S > 1/100 + 1/100 + 1/100 +...+ 1/100 (50 phân số) => S > 1/2.

Vậy S > 1/2

\(S=\left(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}\right)\)

Có: \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

\(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)

=> \(S>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)=> đpcm

Ta có S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\)

\(=\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\right)\)

               25 số hạng                                                    25 số hạng

\(>\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\right)\)

\(=25.\frac{1}{75}+25.\frac{1}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)(ĐPCM)

Vậy S > 1/2

ta có:1/50>1/100

         1/51>1/100

          ...............

          1/99>1/100

=>S>50*1/100

=>S>1/2(đpcm)

1/50>1/100

1/51>1/100

...................

1/99>1/100

=>S>50*1/100(do từ 1/50 đến 1/99 có 50 số hạng)

=>S>1/2

EM có thể tham khảo video này:

https://www.youtube.com/watch?v=fBjsHQKClNA&index=7&list=PLq0mRSDfY0BAMTu98fNHi-Lg_E9BWDYhV

@Miyuki Misaki, @Nguyễn Trúc Giang, @Nguyễn Lê Phước Thịnh, @White Hold

a, Ta có : S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\)

⇔ S = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)\)

⇔ \(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}\right)\)

⇔\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\) ( 99 số hạng)

⇔ S = \(\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

⇔ S = \(\frac{5}{6}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

Mà ta có \(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) < 0

⇔ \(-\)\(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) > 0

Như vậy ta được S > \(\frac{5}{6}\) đpcm

b, \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..+\frac{1}{99}+\frac{1}{100}\) ( 91 số hạng)

Ta có \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};..;\frac{1}{99}>\frac{1}{100}\)

⇒ \(A>\frac{1}{10}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) (90 số hạng 100)

⇒ A \(>\frac{10}{100}+90.\frac{1}{100}\)

⇒ A > \(\frac{10}{100}+\frac{90}{100}\)

⇒ A > \(\frac{100}{100}=1\)

Vậy ...