Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

Đặt : \(B=\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)

\(B=\left(\dfrac{99}{1}+1\right)+\left(\dfrac{98}{2}+1\right)+...+\left(\dfrac{1}{99}+1\right)-99\)

\(B=\dfrac{100}{1}+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}-99\)

\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\left(100-99\right)\)

\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\dfrac{100}{100}\)

\(B=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)

Ta có : \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

Mình làm được một câu thôi, bạn dựa vào làm nha!

Ta có:

\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

Đặt vế đầu là A, vế sau là B.

Vế A:

- Tử:

\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)

\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

Vậy:

\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)

Vế B:

- Tử:

\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)

Vậy:

\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)

Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)

@Tuấn Anh Phan Nguyễn giúp mk!!

+)Đặt A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)

A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\left(1+1+1+...+1\right)\) (99 chữ số 1)

A= \(\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

A= \(\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)

A= \(100.\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{2}+\dfrac{1}{100}\right)\)

⇒ M= \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)

M= \(\dfrac{100.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)

M= 100 (1)

+) Đặt B= \(92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)

B= \(\left(1+1+1+...+1\right)-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\) ( 92 chữ số 1)

B= \(\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)

B= \(\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)

B= \(8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)

⇒ N= \(\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)

N= 8 (2)

Từ (1) và (2)⇒ \(\dfrac{M}{N}\) = \(\dfrac{100}{8}\)= \(\dfrac{25}{2}\)

Vậy \(\dfrac{M}{N}=\dfrac{25}{2}\)