Question

tính\(p=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-......-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)hepl meeeeeeeee

Answer 1

\(P=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(P=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(P=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(P=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(P=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

Xong !