Ta có: 1/12>1/22 ; 1/13> 1/22.....1/21>1/22 
Vậy: 1/12+1/13+...+1/22 > 1/22+1/22+1/22+...+1/22 = 11/22 = 1/2 (có 11 số hạng1/22). 
hay: A>1/2 

a: Ta có

A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)

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

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

⇒ A > 1

vậy A > 1

b: ta có

S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))

⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))

⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)

⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)

⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)

vậy S > \(\dfrac{1}{2}\)

Ta có : \(B=\dfrac{1}{12}>\dfrac{1}{22};\dfrac{1}{13}>\dfrac{1}{22};....;\dfrac{1}{21}>\dfrac{1}{22}\)

Vậy : \(B=\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{22}>\dfrac{1}{22}+\dfrac{1}{22}+\dfrac{1}{22}+...+\dfrac{1}{22}=\dfrac{11}{22}=\dfrac{1}{2}\)

( Có 11 số hạng \(\dfrac{1}{2}\))

Hay B \(>\dfrac{1}{2}\)

Lời giải:

a, Ta có: \(A=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+...+\frac{1}{22}>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{1}{22}.11=\frac{11}{22}=\frac{1}{2}\)

Vậy: \(A>\frac{1}{2}\)

b, Ta có: \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\)

\(=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Mà: \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\text{​​}\text{​​}\text{​​}>\left(\frac{1}{50}+...+\frac{1}{50}+\frac{1}{50}\right)+\left(\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\right)\)

=> \(B\text{​​}\text{​​}\text{​​}>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41+25}{50}=\frac{33}{25}>1\)

Vậy: \(B>1\)

c, Ta có: \(C=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{5}+\frac{1}{6}+\left(\frac{1}{7}+...+\frac{1}{7}+\frac{1}{7}\right)=\frac{11}{30}+11.\frac{1}{7}=\frac{407}{210}< \frac{420}{210}=2\)

Vậy: \(C< 2\)

Chúc bạn học tốt!Tick cho mình nhé!

\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{22}>\frac{1}{2}\)

Ta có: \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}>\frac{1}{20}\) (vì từng phân số lớn hơn \(\frac{1}{20}\))

\(\Rightarrow\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

Mà \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{2}\)

\(\Rightarrow\) \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{22}>\frac{1}{2}\)

Chúc bn học tốt