Đặt $A= \frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{x.(x+3)}$
⇔ $3A= \frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{x.(x+3)}$
⇔ $3A= \frac{1}{5}-\frac{1}{x+3}$
⇔ $A= \frac{x+3-5}{5.(x+3)}:3$
⇔ $A= \frac{x-2}{15.(x+3)}$
⇒ $\frac{x-2}{15.(x+3)}= \frac{1}{6}$
⇔ $6x-12= 15x+45$
⇔ $9x= -57$
⇔ $x=\frac{-19}{3}$
⇒ Không có x thỏa mãn