1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3 
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4 
=> A>1

giangtuantai ơi ! Bạn vẫn đi copy à ?

1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+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 
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1

Chỉ cần 30 số hạng đầu đã lớn hơn 1. 
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+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

                         A = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 )          

       A =  1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 ) >  1 / 10 + ( 1 / 100 + 1 / 100 + ... + 1 / 100 )  

          = 1 / 10 + 90 / 100 = 1       

                       Vậy A > 1

1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+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

đúng nhé

\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)

Vậy A > 1

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

 Vì A > 1/91+1/91+...+1/91=1/91*91=1

 Vậy A>1

30 số hạng đầu lớn hơn 1 

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

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

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

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

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

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

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

\(A>1\left(đpcm\right)\)

a>1(đpcm)