Cho M = $\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+…+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{100}}$ ; N = $\frac{92-\fra

Đáp án:

 250%

Giải thích các bước giải:

 M =  $\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+…+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{100}}$ 

 M = $\frac{1+(\frac{1}{99}+1)+(\frac{2}{98}+1)+(\frac{3}{97}+1)+…+(\frac{98}{2}+1)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{100}}$ 

M  = $\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+…+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{100}}$ 

M = $\frac{100.(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+…+\frac{1}{2})}{(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{100})}$  = 100

N  = $\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-…-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+…+\frac{1}{500}}$ 

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

N  = $\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+…+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+…+\frac{1}{500}}$ 

N  = $\frac{8(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+…+\frac{1}{100})}{\frac{1}{5}(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+…+\frac{1}{100})}$ = $\frac{8}{\frac{1}{5}}$ =40

M : N = $\frac{100}{40}$ = 250%