1/5.8+1/8.11+....+1/x(x+3)=1/6
=> 3/5.8 + 3/8.11 +.....+ 3/x(x+3)=1/2
=> 1/5-1/x+3=1/2
=> 1/x+3=1/5-1/2=-3/10=1/-10/3
=>x+3=-10/3=>x=-19/3
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{1}{2}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{x+3}=\frac{-3}{10}\)
\(\Rightarrow\left(-3\right)\left(x+3\right)=10\)
\(\Rightarrow x+3=\frac{-10}{3}\)
\(\Rightarrow x=\frac{-19}{3}\)
Vậy \(x=\frac{-19}{3}\)
