Đặt \(A=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)
\(A=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-+\frac{1}{x+3}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
\(A=\frac{1}{3}\left(\frac{\left(x+3\right)-5}{5\left(x+3\right)}\right)\)
\(A=\frac{\left(x+3\right)-5}{15\left(x+3\right)}\)
1504 không chia hết cho 3;5 nên ta xét tủ :
x + 3 - 5 = 101
x + 3 = 106
x = 103
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3=305\)
VẬY x = 305
