\(B=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

\(=\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+....+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)(có 92 số 1)

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

\(=8:\frac{1}{5}=40\)

\(B\)\(=\)\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}....+\frac{1}{500}}\)

Tham khảo bài làm bn Đàm đi

Hok tốt

 Tử số: 

\(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\)

= \(\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+\left(1-\frac{3}{11}\right)+...+\left(1-\frac{92}{100}\right)\)

= \(\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}\)

= \(8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

Mẫu số: 

\(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\)

= \(\frac{1}{5\times9}+\frac{1}{5\times10}+\frac{1}{5\times11}+...+\frac{1}{5\times100}\)

= \(\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(\Rightarrow\frac{8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}\)

   Bỏ biểu thức \(\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)ở trên và ở dưới (vì biểu thức đó đều có ở trên và ở dưới nên phải gạch bỏ)

\(\Leftrightarrow\frac{8}{\frac{1}{5}}\)= \(8\div\frac{1}{5}=40\)

Vậy N = 40.

dfghfgiu

Công thức:

\(A=3-\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)                        nha bạn   TF Boys

mk nghĩ là nguyễn việt hoàng làm sai rồi!

Đặt: \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(=\frac{1-\left[\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right]}{1-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}\)

\(=\frac{1-\frac{99}{1}}{1-\frac{1}{100}}\)

\(M=\frac{-98}{99}\)

Đặt \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

\(=\frac{92+\left[\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\right]}{1-\left[\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right]}\)

\(=\frac{92+\frac{92}{100}}{1-\frac{1}{500}}\)

\(=\frac{92+\frac{92}{100}}{\frac{499}{500}}\)

Tự làm tiếp đi!