Question

tìm x biết

\(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{x.\left(x-3\right)}=\frac{101}{1540}\)

Answer 1

Mk sửa lại đề nha:

                \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x-2\right)}=\frac{101}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x-2}-\frac{1}{x}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{3}-\frac{1}{x}=\frac{101}{1540}:\frac{1}{2}=\frac{101}{770}\)

\(\Leftrightarrow\)\(\frac{1}{x}=\frac{1}{3}-\frac{101}{770}=\frac{467}{2310}\)

\(\Leftrightarrow\)\(x=\frac{2310}{467}\)

P/S:  Tham khảo nhé!!!