Question

tính :

\(\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+...+\frac{1119}{1120}\)

Answer 1

Đặt A=\(\frac{9}{10}+\frac{39}{40}+...+\frac{1119}{1120}\)

=>A=\(\frac{10-1}{10}+\frac{40-1}{40}+...+\frac{1120-1}{1120}\)

=>A=\(1-\frac{1}{10}+1-\frac{1}{40}+...+1-\frac{1}{1120}\)

=>A=\(11-\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\right)\)

Đặt B=\(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\)

=>3B=\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{1120}\)

=>3B=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)

=>3B=\(\frac{33}{70}\)

=>B=\(\frac{11}{70}\)

=>A=11-\(\frac{11}{70}\)

=>A=\(\frac{759}{70}\)