\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{999}-\frac{1}{1000}+1\)

\(\frac{1}{1}-\frac{1}{1000}+1\)

\(\frac{999}{1000}+1\)

\(\frac{1999}{1000}\)

theo minh bang 1

\(Tacó:\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

bang 1999/1000

1/1.2+1/2.3+1/3.4+...+1/999.1000+1

=1-1/2+1/2-1/3+1/3-1/4+...+1/998-1/999+1/999-1/1000+1

=1-1/1000+1

=999/1000+1

=1999/1000

Chuẩn ko cần chỉnh

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)

=1/1-1/1000=999/1000

1 và 999/1000

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.100}+1\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{100}\)+1

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

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

=\(\frac{199}{100}\)

sorry mk lộn bài này mới đúng :

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)+1

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

=\(1-\frac{1}{1000}+1\)

=\(\frac{999}{1000}+1\)

=\(\frac{1999}{1000}\)

1999/1000

tớ gặp bài này rồi, nhớ k nhé

11/2 bn nhé

mình làm trên violympic rui ra 11/2

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)

\(=\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{10}{9.10}-\frac{9}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{5}{10}-\frac{1}{10}\)

\(=\frac{2}{5}\)

1999/1000

tớ gặp bài này rồi, k nhé

đúng thế còn cách làm tớ biết rồi!

noi cach lam di

Đặt A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{999.1000}+1\)

=> A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{999}-\frac{1}{1000}+1\)

=> A = \(1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

2 nha bạn