giá trị của biểu thức (1-1/2) .(1-1/2) .(1-1/3).(1-1/4) . (1-1/5) . ..... .(1-1/2003).(1-1/2004)
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BQ
1
NT
1
7 tháng 8 2016
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(B=\frac{1}{2004}\)
KV
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) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
TV
1
TN
21 tháng 6 2016
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot....\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot....\cdot2003}{2\cdot3\cdot....\cdot2004}\)
\(=\frac{1}{2004}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right).....\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot.....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1.1.2.3.4.....2002.2003}{2.2.3.4.5....2003.2004}=\frac{1.1}{2.2004}=\frac{1}{4008}\)
nham { 1-1/2} . { 1- 1/3},,,,,,,,,,,,,