\(B=\frac{1}{12}+\frac{1}{13}+...+\frac{1}{22}\)có 11 số hạng

Ta có: \(\frac{1}{12}>\frac{1}{22}\)

           \(\frac{1}{13}>\frac{1}{22}\)

              .............

           \(\frac{1}{22}=\frac{1}{22}\)

\(\Rightarrow B>\left(\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}\right)=\frac{11}{22}=\frac{1}{2}\)

\(A=\frac{10}{27}+\frac{9}{16}\frac{11}{34}\)

Ta có: \(\frac{10}{27}< >\backslash\left(\frac{9}{16}< >\backslash\left(\frac{11}{34}< >Nên\backslash\left(A< >b\right)\right)\right)\backslash\left(B=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{22}\right)\)

\(B>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=11.\frac{1}{22}=\frac{1}{2}\)

Nên \(B>\frac{1}{2}\)

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: \(C< 2\)

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\(\dfrac{1}{12}>\dfrac{1}{22};\dfrac{1}{13}>\dfrac{1}{22};...;\dfrac{1}{21}>\dfrac{1}{22};\dfrac{1}{22}=\dfrac{1}{22}\)

\(\Rightarrow\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{22}>\dfrac{1}{22}.11\) (do A có 11 số hạng)

\(\Leftrightarrow A>\dfrac{11}{22}=\dfrac{1}{2}\) ( đpcm)

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