Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}=\frac{1}{2004}\)
2)
đặt a= 1+2-3-4+5+6-........+2002-2003-2004+2005+2006
Biểu thức a có (2006-1)/1+1=2006(số hạng)
Nhóm 4 số hạng vào một nhóm ta có 2006 / 4= 501 dư 2 số hạng để ra một số đầu và một số cuối
a= 1+(2-3-4+5)+(6-7-8+9)-.........+(2002-2003-2004+2005) + 2006
a=1+0+0+......+0+2006
a=1+2006
a=2007
vậy a = 2007
a/
$A-3=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}-3$
$=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})-3$
$=\frac{2}{2003}-\frac{1}{2004}-\frac{1}{2005}$
$=(\frac{1}{2003}-\frac{1}{2004})+(\frac{1}{2003}-\frac{1}{2005})$
$>0+0=0$
$\Rightarrow A>3$
b/
$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}$
$=1-\frac{1}{2015}<1$
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2002}{2003}.\frac{2003}{2004}\)
\(B=\frac{1.2.3.....2002.2003}{2.3.4.....2003.2004}\)
\(B=\frac{1}{2004}\)
B = (1 - 1/2).(1 - 1/3).(1 - 1/4).(1 - 1/5)...(1 - 1/2003).(1 - 1/2004)
B = 1/2 . 2/3 . 3/4 . 4/5 .... 2002/2003 . 2003/2004
B = 1.2.3.4...2002.2003/2.3.4.5...2003.2004
B = 1/2004
A=1/2+1/2^2+...+1/2^2004
=>2A=1+1/2+...+1/2^2003
=>A=1-1/2^2004=(2^2004-1)/2^2004
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) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
B=1/2+1/2^2+1/2^3+...+1/2^2004
2B=1+1/2+1/2^2+...+1/2^2003
2B-B=(1+1/2+1/2^2+...+1/2^2003)-(1/2+1/2^2+1/2^3+...+1/2004)
2B-B=1+1/2+1/2^2+...+1/2^2003-1/2-1/2^2-1/2^3-...-1/2^2004 (TỐI GIẢN CÁC PHÂN SỐ GIỐNG NHAU)
B=1-1/2^2004