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

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

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

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

\(A>1\left(đpcm\right)\)

a>1(đpcm)

Ta có : 

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

​Vậy :C>1

sao lại chọn người không tự làm mà đi copy ?      

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

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 > \(\dfrac{1}{2}\)

1/10>1/40;1/11>1/40;...;1/40=1/40

=>1/10+1/11+...+1/40>1/40+1/40+...+1/=1/40 .31=31/40>1/2                              (1)

1/41>1/100;1/42>1/100;....1/100=1/100

=>1/41+1/42+...+1/100>1/100+1/100+...+1/100=1/100 .60=3/5>1/2                     (2)

Từ (1) và (2) =>A>1/2+1/2=1

Vậy A>1(đpcm)

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)

A = 1/10 + 90/100 = 1

vậy A > 1

 Vì A > 1/91+1/91+...+1/91=1/91*91=1

 Vậy A>1

30 số hạng đầu lớn hơn 1 

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

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

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

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