đề là gì vậy bạn

Lời giải:

$A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{26-25}{25.26}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{25}-\frac{1}{26}$

$=1-\frac{1}{26}< 1$ (đpcm)

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

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