Tính: $\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{109} }{\frac{100}{1}+\frac{99}{2}+\frac{98}{3}+…+\frac{1}{100}}$ Plz, mai nộp rùi ak.

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1. $\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{109}}{\frac{100}{1}+\frac{99}{2}+\frac{98}{3}+…+\frac{1}{100}}$ là sai. Phải là $\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{101}}{\frac{100}{1}+\frac{99}{2}+\frac{98}{3}+…+\frac{1}{100}}$

   Đặt B=$\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{101}}{\frac{100}{1}+\frac{99}{2}+\frac{98}{3}+…+\frac{1}{100}}$

   Xét: ${\frac{100}{1}+\frac{99}{2}+\frac{98}{3}+…+\frac{1}{100}}$

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

   =$\frac{101}{2}$ + $\frac{101}{3}$ + …+$\frac{101}{100}$ + $\frac{101}{101}$ 

   =101.($\frac{1}{2}$+$\frac{1}{3}$ +…+$\frac{1}{100}$ + $\frac{1}{101}$  )

   ⇒B=$\frac{\frac{1}{2}+\frac{1}{3} +…+\frac{1}{100} + \frac{1}{101}}{101.(\frac{1}{2}+\frac{1}{3}+…+\frac{1}{100} + \frac{1}{101})}$ 

   ⇒B=$\frac{1}{101}$ 

   Vậy B=$\frac{1}{101}$ 

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