\(A=\frac{3^2-1}{5^2-1}:\frac{9^2-1}{7^2-1}:\frac{13^2-1}{11^2-1}:...:\frac{55^2-1}{53^2-1}\)

    \(=\frac{\left(3-1\right)\left(3+1\right)}{\left(5-1\right)\left(5+1\right)}:\frac{\left(9-1\right)\left(9+1\right)}{\left(7-1\right)\left(7+1\right)}:\frac{\left(13-1\right)\left(13+1\right)}{\left(11-1\right)\left(11+1\right)}:...:\frac{\left(55-1\right)\left(55+1\right)}{\left(53-1\right)\left(53+1\right)}\)

    \(=\frac{2.4}{4.6}:\frac{8.10}{6.8}:\frac{12.14}{10.12}:...:\frac{54.56}{52.54}\)

    \(=\frac{2.4.6.8.10.12......52.54}{4.6.8.10.12.....54.56}\)

    \(=\frac{2}{56}\)

   \(=\frac{1}{28}\)

Ta có :

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

cảm ơn bạn nha

sfdsa

VÌ 1/1.1/3.......1/99=2/51.2/52.........2/100

VÀ   2/51.2/52.....2/100=1/1.1/3.......1/99

SUY RA BẰNG NHAU

  A= 1 + 5 + 5² + 5 ³ + ... + 5⁸⁰⁰ 

5A=       5 + 5²  + 5³ + .... +5 ⁸⁰⁰ + 5⁸⁰¹  

5A - A = 5⁸⁰¹  - 1 

4a = 5⁸⁰¹ - 1 

    5⁸⁰¹ - 1 +1 = 5ⁿ

⇒  5⁸⁰¹ = 5ⁿ ⇒ n = 801

Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)