em lp 5 nen ko biet!

\(\frac{50}{111}>\frac{1}{4};\frac{50}{112}>\frac{1}{4};\frac{50}{113}>\frac{1}{4};\frac{50}{114}>\frac{1}{4}\)

\(A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}>\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)(1)

\(\frac{50}{111}< \frac{1}{2};\frac{50}{112}< \frac{1}{2};\frac{50}{113}< \frac{1}{2};\frac{50}{114}< \frac{1}{2}\)

\(\Rightarrow A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}< \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)(2)

từ (1) và (2) \(\Rightarrow1< A< 2\)

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 \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)

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

\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(< \frac{1}{26}+\frac{1}{26}+\frac{1}{26}+...+\frac{1}{26}+\frac{1}{26}\)
\(=\frac{25}{26}< 1\)(sai với đề bài)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\Rightarrow A=B\text{(đpcm)}\)

Ta cos ..............

suy ra A=B

Giải:

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

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

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

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

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

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