Question

tìm x biết \(\dfrac{1}{x}\)-\(\dfrac{1}{9999}\)=\(\dfrac{1}{\begin{matrix}1&\times&3\end{matrix}}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{97\times99}\)

Answer 1

Đặt

\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)

\(S=\dfrac{49}{99}\)

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

\(\Rightarrow\dfrac{1}{x}=\dfrac{50}{101}\Leftrightarrow50x=101\Leftrightarrow x=\dfrac{101}{50}\)

Answer 2

Đặt \(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(M=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(M=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)

\(M=\dfrac{1}{2}\cdot\dfrac{98}{99}\)

\(M=\dfrac{49}{99}\)