Question

Tìm \(x\) biết :

\(\frac{1}{5.8}.\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

Answer 1

\(\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{x\times\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{x\times\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}\div\frac{1}{3}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}\times3\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(x+3=308\)

\(x=308-3\)

x = 305

Chúc bạn học tốt ^^

Answer 2

\(\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\frac{1}{3}.\frac{1}{5}-\frac{1}{3}-\frac{1}{x+3}=\frac{101}{1540}\)

\(\frac{1}{15}-\frac{1}{x+3}=\frac{101}{1540}\)

\(\frac{1}{x+3}=\frac{1}{15}-\frac{101}{1540}\)

\(\frac{1}{x+3}=\frac{1}{924}\)

=> x = 924 -3

=> x = 921