Bài giải :

$\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left( x+1 \right)}=\frac{2}{9}$

$\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x\left( x+1 \right)}=\frac{2}{9}$

$2\left( \frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1} \right)=\frac{2}{9}$

$2\left( \frac{1}{6}-\frac{1}{x+1} \right)=\frac{2}{9}$

Vậy x=17.