Ta có : \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{1599}{1600}\)

\(=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{1600}\right)\)

Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)

\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{1601}\right)\)

Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{6}>\frac{1}{7};...;\frac{1}{1600}>\frac{1}{1601}\)

\(\Rightarrow1-\frac{1}{2}< 1-\frac{1}{3};1-\frac{1}{4}< 1-\frac{1}{5};1-\frac{1}{6}< 1-\frac{1}{7};...;1-\frac{1}{1600}< 1-\frac{1}{1601}\)

\(\Rightarrow A< B\)

hay A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)

Vậy A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\).

Ta luôn có: 

\(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{7}< \frac{6}{7}\)

\(........\)

\(\frac{1599}{1600}< \frac{1600}{1601}\)

Từ trên: \(\Rightarrow A=\frac{1}{2}.\frac{3}{4}....\frac{1599}{1600}\left(1\right)\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}...\frac{1599}{1600}< \frac{2}{3}.\frac{4}{5}....\frac{1600}{1601}\left(2\right)\)

Từ: \(\left(1\right)\left(2\right)\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\left(đpcm\right)\)

bạn ơi trả lời họ mình với

là sao

vào trả lời hộ

\(=\frac{2.4.3.5.4.6.5.7....39.41}{3.4.5....40.3.4.5....40}=\frac{\left(2.3.4.....39\right).\left(4.5.6.....41\right)}{\left(3.4.5......40\right).\left(3.4.5......40\right)}=\frac{2.3.4.....39}{3.4.5....40}.\frac{4.5.6.....41}{3.4.5.....40}=\frac{2}{40}.\frac{41}{3}=\frac{41}{60}\)

15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9

Sửa đề: 

A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007

=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007

=-1+1+1/2007=1/2007

16:

\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)

=1/64

17:

=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7

=-6/7