Ta có:
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}\)
\(=x.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(=x.\left(\frac{7}{14}-\frac{1}{14}\right)=x.\frac{3}{7}=\frac{1}{21}\)
\(\Rightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}.\frac{7}{3}=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)