Ta có:
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x.\left(x+4\right)}=\frac{5}{63}\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{x.\left(x+4\right)}\right)=\frac{5}{63}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{x}-\frac{1}{x+4}\right)=\frac{5}{63}:\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{20}{63}\Leftrightarrow\frac{1}{x+4}=\frac{1}{63}\Leftrightarrow x=63-4=59\)