\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}.\frac{1}{2}\)
\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}=\frac{1}{9}\)
=>x+1=9
=>x=8