Nhận xét thấy:

\(\dfrac{1}{1.2}\)= 1-\(\dfrac{1}{2}\); \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\);...

Ta có

A= 1-\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

A= 1- \(\dfrac{1}{6}\)

A= \(\dfrac{5}{6}\)

Vậy A= \(\dfrac{5}{6}\)