Ta chứng minh được :$\frac{1}{a\left( a+2 \right)}=\frac{1}{2}\left( \frac{1}{a}-\frac{1}{a+2} \right)$

Áp dụng vào biểu thức có:

$\frac{1}{2}\left( \frac{1}{1}-\frac{1}{3} \right)+\frac{1}{2}\left( \frac{1}{3}-\frac{1}{5} \right)+\frac{1}{2}\left( \frac{1}{5}-\frac{1}{7} \right)+...+\frac{1}{2}\left( \frac{1}{x}-\frac{1}{x+2} \right)$

$=\frac{1}{2}.\frac{1}{1}-\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{5}+\frac{1}{2}.\frac{1}{5}-\frac{1}{2}.\frac{1}{7}+...-\frac{1}{2}.\frac{1}{x}+\frac{1}{2}.\frac{1}{x}-\frac{1}{2}.\frac{1}{x+2}$

$=\frac{1}{2}-\frac{1}{2}.\frac{1}{x+2}=\frac{8}{17}$

$\to 1-\frac{1}{x+2}=\frac{16}{17}$

$\to \frac{1}{x+2}=\frac{1}{17}$

$\to x+2=17$

 Suy ra x = 15