Ta có :

\(\frac{50}{111}>\frac{50}{200}\)

\(\frac{50}{112}>\frac{50}{200}\)

\(\frac{50}{113}>\frac{50}{200}\)

\(\frac{50}{114}>\frac{50}{200}\)

\(\Rightarrow A>\frac{50}{200}+\frac{50}{200}+\frac{50}{200}+\frac{50}{200}\)hay \(A>\frac{50}{200}.4\left(1\right)\)

Mặt khác :

\(\frac{50}{111}< \frac{50}{100}\)

\(\frac{50}{112}< \frac{50}{100}\)

\(\frac{50}{113}< \frac{50}{100}\)

\(\frac{50}{114}< \frac{50}{100}\)

\(\Rightarrow A< \frac{50}{100}+\frac{50}{100}+\frac{50}{100}+\frac{50}{100}\)hay \(A< \frac{50}{100}.4\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)

50/111 < 50/100

50/112 < 50/100

50/113 < 50/100 

50/114 < 50/100

=> A < 200/100 => A < 2

50/111 > 50/200

50/112 > 50/200

50/113 > 50/200

50/114 > 50/200

=> A > 200/200 => A > 1

Vậy 1 < A < 2

AI THẤY OK ỦNG HỘ NHÉ 

Ta có :

\(A=\dfrac{50}{111}+\dfrac{50}{112}+\dfrac{50}{113}+\dfrac{50}{114}\)

Ta thấy :

\(\dfrac{50}{111}>\dfrac{50}{200}\)

\(\dfrac{50}{112}>\dfrac{50}{200}\)

\(\dfrac{50}{113}>\dfrac{50}{200}\)

\(\dfrac{50}{114}>\dfrac{50}{200}\)

\(\Rightarrow A>\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}\)

\(\Rightarrow A>\dfrac{50}{200}.4=1\) \(\left(1\right)\)

Mặt khác :

\(\dfrac{50}{111}< \dfrac{50}{100}\)

\(\dfrac{50}{112}< \dfrac{50}{100}\)

\(\dfrac{50}{113}< \dfrac{50}{100}\)

\(\dfrac{50}{114}< \dfrac{50}{100}\)

\(\Rightarrow A< \dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}\)

\(\Rightarrow A< \dfrac{50}{100}.4=2\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Rightarrow1< A< 2\rightarrowđpcm\)

A<50/100+50/100+50/100+50/100=4.50/100=2

=>A<2

A>4.50/150=4/3+1+1/3>1

=>dccm

Ta biến đổi vế phải : 

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\\ \)\(\\ =\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}\)

Vậy \(\frac{1}{26}+\frac{1}{27}+.....+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)

Ta có

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)

=> \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\) ( đpcm )

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{60}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)