C>1   vì c>1

a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)

Vậy A > 1/2

b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Vậy B > 1/2

c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)

Vậy C > 1

a: Ta có: \(A=5^0+5^1+\cdots+5^{2011}\)

=>5A=\(5+5^2+\cdots+5^{2012}\)

=>5A-A=\(5+5^2+\ldots+5^{2012}-1-5-5^2-\cdots-5^{2011}\)

=>4A=\(5^{2012}-1\)

=>4A+1=\(5^{2012}\)

=>4A+1 là lũy thừa cơ số 5

b: \(4A+1=5^{x}\)

=>\(5^{x}=5^{2012}\)

=>x=2012

c: \(A=5^0+5^1+5^2+5^3+\cdots+5^{2010}+5^{2011}\)

\(=\left(1+5\right)+5^2\left(1+5\right)+\cdots+5^{2010}\left(1+5\right)\)

\(=6\left(1+5^2+\cdots+5^{2010}\right)\)

=>A⋮6

d: \(A=5^0+5^1+5^2+5^3+\cdots+5^{2010}+5^{2011}\)

\(=\left(1+5\right)+5^2\left(1+5+5^2\right)+5^5\left(1+5+5^2\right)+\cdots+5^{2009}\left(1+5+5^2\right)\)

=6+\(31\left(5^2+5^5+\cdots+5^{2009}\right)\)

=>A chia 31 dư 6

Ta ó: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};....;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow S>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\left(50so\right)=\frac{50}{100}=\frac{1}{2}\)

Vậy...

Ta có :

Tất cả các số hạng của tổng đều lớn hơn \(\frac{1}{100}\), mà tổng có 50 số hạng

=> S > \(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)( có 50 số 1/100 )

=> S > \(\frac{50}{100}\)= \(\frac{1}{2}\)

Vậy S > 1/2

@Miyuki Misaki, @Nguyễn Trúc Giang, @Nguyễn Lê Phước Thịnh, @White Hold

Ta có: \(\frac{1}{50}\)>\(\frac{1}{100}\)

\(\frac{1}{51}\)>\(\frac{1}{100}\)

\(\frac{1}{52}\)>\(\frac{1}{100}\)

..................

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

=>\(\frac{1}{50}\)+\(\frac{1}{51}\)+.............+\(\frac{1}{99}\)>\(\frac{1}{100}\).50=\(\frac{1}{2}\)(50 là số số hạng  của S nha)

=>S>\(\frac{1}{2}\)

a: \(A=\frac{1}{1\cdot1}+\frac{1}{2\cdot3}+\frac{1}{3\cdot5}+\frac{1}{3\cdot7}+\cdots+\frac{1}{50\cdot99}\)

\(=2\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\right)\)

\(=2\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left\lbrack\left(1+\frac12+\frac13+\frac14+\cdots+\frac{1}{100}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\right\rbrack\)

\(=2\left\lbrack1+\frac12+\frac13+\cdots+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\right\rbrack\)

\(=2\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)\)

b: Ta có: \(\frac{1}{51}>\frac{1}{75};\frac{1}{52}>\frac{1}{75};\ldots;\frac{1}{75}=\frac{1}{75}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (1)

Ta có: \(\frac{1}{76}>\frac{1}{100};\frac{1}{77}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}=\frac{25}{100}=\frac14\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)>\frac13+\frac14=\frac{7}{12}\)

=>\(A>\frac{7}{12}\cdot2=\frac76\)