\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{18}\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{18}:\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}\)
<=>x+3=30
<=>x=27
Vậy x=27
