Question

[Dương Đình Hưởng]

Tính:

D= \(\frac{1}{100.99}\)- \(\frac{1}{99.98}\)- \(\frac{1}{98.97}\)-...- \(\frac{1}{3.2}\)- \(\frac{1}{2.1}\).

Answer 1

[kudo shinichi]

\(D=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(D=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(D=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(D=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)

\(D=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)

b tự làm nốt nhé

Answer 2

[Lê Quang Tuấn Kiệt]

D=1100.99 −199.98 −198.97 −...−13.2 −12.1 

D=199.100 −(11.2 +12.3 +...+197.98 +198.99 )

D=199 −1100 −(1−12 +12 −13 +...+198 −199 )

D=199 −1100 −(1−199 )

D=199 −1100 −1+199