1/1000

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

\(A=\frac{1.2.3...999}{2.3.4...1000}\)

\(A=\frac{1}{1000}\)

vậy A = B

đề là 1/2x3/4x5/6x....x999/1000 ao bạn hiếu lại ghi một nẻo khác z

Hiếu sai rồ đáp án đúng là A>B

????

Đề thiếu. Bạn xem lại đề.

help me , pls

Giải:

\(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}\)

Đặt \(B=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{10000}{10001}\)

Do \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

Nên \(C< B\) Mà \(\left\{{}\begin{matrix}C>0\\B>0\end{matrix}\right.\)

\(\Rightarrow C^2< C.B=\left(\dfrac{1}{2}.\dfrac{3}{4}...\dfrac{9999}{10000}\right)\)\(\left(\dfrac{2}{3}.\dfrac{4}{5}...\dfrac{10000}{10001}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{9999}{10000}.\dfrac{10000}{10001}\)

\(=\dfrac{1.2.3.4.5.6...9999.10000}{2.3.4.5.6.7....10000.10001}\)

\(=\dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(\dfrac{1}{100}\right)^2\)

\(\Rightarrow C^2< \left(\dfrac{1}{100}\right)^2\Leftrightarrow C< \dfrac{1}{100}\)

Vậy \(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\) (Đpcm)

\(A=\frac{9999}{32000}=0,31246875...\)

\(\frac{1}{1000}=0,001\Rightarrow0,31246875...>0,001\)

\(\Rightarrow A>\frac{1}{1000}\)