Ta có:
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)
\(=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{1}{100}+\frac{1}{100}-1\)
\(=\frac{1}{50}-\frac{50}{50}\)
\(=-\frac{49}{50}\)
Câu này khó quá ta mình suy nghĩ này giờ mà vẫn chưa ra