Question

Answer 1

nhớ bài này làm rồi mà :v

\(P=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{97\cdot98}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\)

\(P=1-\dfrac{1}{98}\)

\(P=\dfrac{98}{98}-\dfrac{1}{98}\)

\(P=\dfrac{97}{98}\)

Answer 2

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{97}-\dfrac{1}{98}\)

\(=1-\dfrac{1}{98}=\dfrac{97}{98}\)