Question

[Hồ Bình An]

tinh gia tri bieu thuc\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

Answer 1

[Hồ Thu Giang]

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

= \(1-\frac{1}{100}\)

= \(\frac{99}{100}\)

Answer 2

[Đỗ Văn Hoài Tuân]

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)