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 :

A = \(\dfrac{1}{10}\) + \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) +.................+ \(\dfrac{1}{99}\) + \(\dfrac{1}{100}\) ( 91 số hạng)

A = \(\dfrac{1}{10}\) + \(\left(\dfrac{1}{11}+\dfrac{1}{12}+...........+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

Vì \(\dfrac{1}{11}>\dfrac{1}{100}\)

\(\dfrac{1}{12}>\dfrac{1}{100}\)

.................................

\(\dfrac{1}{99}< \dfrac{1}{100}\)

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

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

\(A\) > \(\dfrac{1}{10}+\dfrac{9}{10}\)

=> A > 1

=> đpcm

\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

            \(\frac{13}{12}\)        \(>\)         \(1\)

@Miyuki Misaki, @Nguyễn Trúc Giang, @Nguyễn Lê Phước Thịnh, @White Hold

a, Ta có : S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\)

⇔ S = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)\)

⇔ \(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}\right)\)

⇔\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\) ( 99 số hạng)

⇔ S = \(\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

⇔ S = \(\frac{5}{6}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

Mà ta có \(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) < 0

⇔ \(-\)\(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) > 0

Như vậy ta được S > \(\frac{5}{6}\) đpcm

b, \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..+\frac{1}{99}+\frac{1}{100}\) ( 91 số hạng)

Ta có \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};..;\frac{1}{99}>\frac{1}{100}\)

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

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

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

⇒ A > \(\frac{100}{100}=1\)

Vậy ...

\(S=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{99.100}\)

\(\Rightarrow S=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow S=\frac{1}{10}-\frac{1}{100}\)

\(\Rightarrow S=\frac{99}{100}\)

\(S=\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{99.100}\)

\(=\frac{11-10}{10.11}+\frac{12-11}{11.12}+...+\frac{100-99}{99.100}\)

\(=\frac{11}{10.11}-\frac{10}{10.11}+\frac{12}{11.12}-\frac{11}{11.12}+....+\frac{100}{99.100}-\frac{99}{99.100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}=\frac{9}{100}\)

Thanks bạn Đinh Tuấn Việt nhiều nah!!!!

C>1   vì c>1

a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)

Vậy A > 1/2

b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Vậy B > 1/2

c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)

Vậy C > 1

Tớ đồng ý,bạn làm đúng rồi .......