Question

Tính nhanh:

P = \(\frac{1}{99}\)-\(\frac{1}{99x98}\)-\(\frac{1}{98x97}\)-\(\frac{1}{97x96}\)-...-\(\frac{1}{3x2}\)-\(\frac{1}{2x1}\)

x = dấu nhân.

Gíup mình nhé.

Answer 1

Ta có: \(P=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)

\(=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(=\frac{2}{99}-1=\frac{2}{99}-\frac{99}{99}\)

\(=\frac{-97}{99}\)