Vì A > 1/91+1/91+...+1/91=1/91*91=1

 Vậy A>1

30 số hạng đầu lớn hơn 1 

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

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

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

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

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

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

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

=> 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)

Ta có:A=1/10+1/11+1/12+...+1/99+1/100

           =1/10+(1/11+1/12+...+1/100)

          >1/10+(1/100+1/100+1/100+...+1/100)

           =1/10+90/100=1/10+9/10=1

      Vậy A>1

Mình chúc bạn học tốt

1/10+1/11+……+1/99 > 1/20+1/20+…..+1/20 = 10/20 = 1/2

1/20+1/21+……+1/29 > 1/20+1/30+…..+1/30 = 10/30 = 1/3

1/30+1/31+……+1/39 > 1/40+1/40+…..+1/40 = 10/40= 1/4 

=> 1/10 + 1/11 +...+ 1/39 > 1/2 + 1/3 + 1/4 = 13/12 > 1

Vậy A > 1

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)

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