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

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

\(A=\frac{1}{10}+\frac{99}{100}=1\)

=> A > 1

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

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

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

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

\(\Rightarrow A>1\)

Ta thấy:1/10;1/11;1/12;1/13;...;1/99>1/100

=)1/10+1/11+1/12+1/13+...+1/100>1/100+1/100+1/100+1/100..+1/100=1/100.100=1

Vậy A>1

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

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

\(=\frac{1}{10}+\frac{90}{100}>1\)

\(A>1\left(đpcm\right)\)

a>1(đpcm)

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

    \(>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41}{50}+\frac{50}{100}=\frac{33}{25}=1\frac{8}{25}>1\)

Ta thấy rằng mỗi số hạng trong tổng đều lớn hơn hoặc bằng \(\frac{1}{100}\)

=> \(C>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}x100=1\)

=> C>1 (Đpcm)

ta có : \(\frac{1}{10}>\frac{1}{100}\)

          \(\frac{1}{11}>\frac{1}{100}\)

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

             \(..............\)

          \(\frac{1}{99}>\frac{1}{100}\)

         \(\frac{1}{100}=\frac{1}{100}\)

cộng vế với vế ta được :

\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{91}{100}>1\)

hiệp sai rùi

vầy nè : ta tách a thành 2 nhóm

nhom1 tu 1/10 den 1/50 ta dat =b

ta có :b=1/10+1/11+1/12+.......1/50

          b>41/50 (vi 1/10 >1/50;1/11>1/50;....1/50=1/50)

nhóm 2 =c làm tương tự >50/100

a= b+c>50/100+41/50=33/25>25/25=1 

mình ko giải chi tiết đâu bạn

Ta có: A = \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{10}>\frac{1}{100};\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};....;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow A>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)

\(\Rightarrow A>\frac{1}{10}+\frac{90}{100}=\frac{100}{100}=1\)                 (ĐPCM)

A không thể lớn hơn 1 được

Ta có:

\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{40}{50}=\frac{4}{5}\)

\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Từ đây ta suy ra 

A > \(\frac{4}{5}+\frac{1}{2}+\frac{1}{100}=1,31>1\)  

Hình như có lần đã c/m A>6/5

 Vì A > 1/91+1/91+...+1/91=1/91*91=1

 Vậy A>1

30 số hạng đầu lớn hơn 1 

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

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

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

=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)

Đặt A = B + \(\frac{1}{10}\) Ta thấy B có 90 số hạng và 1/100 < 1/11 ; 1/100 < 1/12 .....

Giả sử cả 90 số hạng đều là 1/100 ta có B > 90.(1/100) = 90/100

=> A > 1/10 + 90/100 => A>1

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

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

Vậy A>1

gỉa sử a là 1/ 10 thì b >1

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)

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

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

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

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) 

Cảm ơn bạn Uyên nhiều nha!

^_^^_^^_^