Question

tìm x : \(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

Answer 1

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)

\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)

\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)

\(\frac{1}{x}=\frac{50}{101}\)

\(x=1:\frac{50}{101}\)

\(x=\frac{101}{50}\)

Vậy \(x=\frac{101}{50}\)