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13 tháng 8 2017

Đặt \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2004}\right)\)

\(A=\left(\dfrac{2}{2}-\dfrac{1}{2}\right).\left(\dfrac{3}{3}-\dfrac{1}{3}\right)....\left(\dfrac{2003}{2003}-\dfrac{1}{2003}\right).\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)

\(A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(A=\dfrac{1.2.3...2002.2003}{2.3....2003.2004}\)

\(A=\dfrac{1}{2004}\)

13 tháng 8 2017

\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)..............\left(1-\dfrac{1}{2004}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.............\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

13 tháng 8 2017

Giúp mình nhanh nha! Mình sẽ thick người đó

11 tháng 8 2016

\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}=\frac{1x2x3x...x2002x2003}{2x3x4x...x2003x2004}=\frac{1}{2004}\)

11 tháng 8 2016

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)

TH
Thầy Hùng Olm
Manager VIP
14 tháng 7 2023

\(B=\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{2003}{2004}\)

\(B=\dfrac{1}{2004}\)

14 tháng 7 2023

\(B=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times....\left(1-\dfrac{1}{2003}\right)\times\left(1-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\times\dfrac{2}{3}\times....\times\dfrac{2002}{2003}\times\dfrac{2003}{2004}\) 

\(B=\dfrac{1}{2004}\)

14 tháng 6 2023

a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{10}{3}\)

\(=\dfrac{3}{5}\)

14 tháng 6 2023

ai trả lời đúng thì mình sẽ tick cho

7 tháng 7 2015

A=1/2 x 2/3 x 3/4 x ... x 2002/2003 x 2003/2004

=1/2004

XEM LẠI GIÙM MÌNH NHA!

23 tháng 7 2018

bo ko bt

18 tháng 8 2023

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right).\)

\(\Rightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(\Rightarrow A=\dfrac{1}{2004}\)

HQ
Hà Quang Minh
Giáo viên
18 tháng 8 2023

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2004}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}...\dfrac{2003}{2004}\\ =\dfrac{1}{2004}\)