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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right)\)\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

\(A=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{35}\right)+\left(\frac{1}{36}+...+\frac{1}{50}\right)>\frac{1}{35}.10+\frac{1}{50}.15=\frac{41}{70}>\frac{7}{12}\)

\(A< \frac{10}{26}+\frac{15}{36}< \frac{5}{6}\) Vậy ....

\(A=\frac{1}{2}+\frac{1}{12}+...+\frac{1}{9900}>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)-\frac{1}{100}<\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{5}{6}\)

=> điều phải c/m nha

Ta có: A=1/1.2+1/3.4+1/5.6+...+1/99.100

             =1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100

             =1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-2(1/2+1/4+1/6+...+1/100)

             =1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-(1+1/2+1/3+1/4+...+1/50)

             =1/26+1/27+1/28+...+1/100)

Do đó A=(1/51+1/52+...+1/75)+(1/76+1/77+...+1/100)

Ta có 1/51>1/52>...>1/75 và 1/76>1/77>...>1/100 nên

A>1/75.25+1/100.25=1/3+1/4=7/12

A<1/51.25+1/76.25<1/50.25+1/75.25=1/2+1/3=5/6

Vậy nên 7/12<A<5/6

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{100}\)

Ta có A =1/1.2+1/3.4+1/5.6+...+1/99.100

=(1/1.2+1/3.4)+(1/5.6+...+1/99.100)

=7/12+(1/5.6+...+1/99.100)>7/12(1)

A=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100

=(1+1/3+1/5+...+1/99)-(1/2+1/4+..+1/100)

=(1+1/2+1/3+1/4+..+1/99+1/100)-2(1/2+1/4+....+1/100)    ( Cộng thêm cả 2 vế với 1/2+1/4+..+1/100)

=(1+1/2+1/3+..+1/100)-(1+1/2+..+1/50)

=1/51+1/52+..+1/100

Dãy số trên có 50 số hang 50 chia hết cho 10 nên ta nhóm 10 số vào 1 nhóm

A=(1/51+1/52+..+1/60)+(1/61+1/62+..+1/70)+(1/71+1/72+..+1/80)+(1/81+..+1/90)+(1/91+..+1/100)

<1/50.10+1/60.10+1/70.10+1/80.10+1/90.10=1/5+1/6+1/7+1/8+1/9<1/5+1/6+1/7.3=167/210<175/210=5/6

=>A<5/6(2)

từ 1 và 2 => đpcm