\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{89\cdot90}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{89}-\dfrac{1}{90}\\ =\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{22}{45}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{89.90}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{89}-\dfrac{1}{90}\\ =\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{89}-\dfrac{1}{89}\right)-\dfrac{1}{90}\\ =\dfrac{1}{2}-0-0-...-0-\dfrac{1}{90}\\ =\dfrac{1}{2}-\dfrac{1}{90}\\ =\dfrac{45}{90}-\dfrac{1}{90}\\ =\dfrac{44}{90}\\ =\dfrac{22}{45}\)
Ta có:
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{89\cdot90}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{89}+\dfrac{1}{90}\)
\(=\dfrac{1}{2}+\dfrac{1}{90}\)
\(=\dfrac{23}{45}\)