Ta có :

\(\frac{1}{10}>\frac{1}{20}\)

\(\frac{1}{11}>\frac{1}{20}\)

\(\frac{1}{12}>\frac{1}{20}\)     \(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+.....+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(1)

.....

\(\frac{1}{19}>\frac{1}{20}\)

Ta có :

\(\frac{1}{20}>\frac{1}{30}\)

\(\frac{1}{21}>\frac{1}{30}\)

\(\frac{1}{22}>\frac{1}{30}\)      \(\Rightarrow\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+....+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)(2) 

........

\(\frac{1}{29}>\frac{1}{30}\)

Ta có :

\(\frac{1}{30}>\frac{1}{40}\)

\(\frac{1}{31}>\frac{1}{40}\)                \(\Rightarrow\frac{1}{30}+\frac{1}{31}+....+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)(3)

.........

\(\frac{1}{39}>\frac{1}{40}\)

Từ 1 , 2 , 3 ,

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

=> ....... > 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 
\(\Rightarrow\)1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1

Câu hỏi của Thăng Phạm - Toán lớp 6 - Học toán với OnlineMath 

Em tham khảo bài bạn làm nhé!

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

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

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

=> A > 1

bài 1:

ta có \(\frac{1}{1!}=1\)

\(\frac{1}{2!}=\frac{1}{1\cdot2}\)

\(\frac{1}{3!}=\frac{1}{1\cdot2\cdot3}=\frac{1}{2\cdot3}\)

bắt đầu từ đây ta giảm mẫu số:

\(\frac{1}{4!}=\frac{1}{1\cdot2\cdot3\cdot4}<\frac{1}{3\cdot4}\)

... tới \(\frac{1}{2012!}=\frac{1}{1\cdot2\cdot\ldots\cdot2011\cdot2012}<\frac{1}{2011\cdot2012}\)

thay vào biểu thức S

=> \(S<1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2011\cdot2012}\)

áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

=> \(S=1+1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2011}-\frac{1}{2012}\)

\(S<2-\frac{1}{2012}\)

mà \(\frac{1}{2012}>0\)

=> \(S<2\)

bài 2:

Ta có công thức: \(\frac{1}{\left(n+1\right)!}=\frac{1}{n!}-\frac{1}{\left(n+1\right)!}\)

=> \(\frac{9}{10!}=\frac{1}{9!}-\frac{1}{10!}\)

\(\frac{10}{11!}=\frac{1}{10!}-\frac{1}{11!}\)

\(\frac{11}{12!}=\frac{1}{11!}-\frac{1}{12!}\)

... tới: \(\frac{99}{100!}=\frac{1}{9!}-\frac{1}{100!}\)

thay vào biểu thức ta gọi biểu thức là A

\(A=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\cdots+\frac{1}{99!}-\frac{1}{100!}\)

A=\(\frac{1}{9!}-\frac{1}{100!}\)

mà \(\frac{1}{100!}>0\Rightarrow\frac{1}{9!}-\frac{1}{100!}<\frac{1}{9!}\)

vậy \(A<\frac{1}{9!}\)

=>1/11*12*13*.............*99*100

=>S<1

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

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

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

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

Làm ơn giải ra luôn hộ

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

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

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

Vậy A > 1 (đpcm)

+)Ta có:\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..........+\frac{1}{99}+\frac{1}{100}\)(có (100-10):1+1=91 số hạng)

\(\Rightarrow A=\left(\frac{1}{10}+\frac{1}{11}+.........+\frac{1}{54}\right)+\frac{1}{55}+\left(\frac{1}{56}+\frac{1}{57}+.............+\frac{1}{100}\right)>\)

\(\left(\frac{1}{54}+\frac{1}{54}+........+\frac{1}{54}\right)+\frac{1}{55}+\left(\frac{1}{100}+\frac{1}{100}+........+\frac{1}{100}\right)\)

\(\Rightarrow A>\frac{45}{54}+\frac{1}{55}+\frac{45}{100}=\frac{5}{6}+\frac{1}{55}+\frac{9}{20}=\frac{5}{6}+\frac{9}{20}+\frac{1}{55}=\frac{50}{60}+\frac{27}{60}+\frac{1}{55}\)\(=\frac{77}{60}+\frac{1}{55}>1\)(vì \(\frac{77}{60}>1\))

\(\Rightarrow A>1\)(ĐPCM)
Chúc bn học tốt

thanks nhìu!

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