Trả lời giúp mk nhé KHẨN CẤP!
Tính:
A=(1 - 1/2)×(1 - 1/3)×(1 - 1/4)×...×(1 - 1/2004)
Thanks
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Đặt A = 1/3 + 1/32 + 1/33 + ... + 1/32005
3A = 1 + 1/3 + 1/32 + ... + 1/32004
3A - A = (1 + 1/3 + 1/32 + ... + 1/32004) - (1/3 + 1/32 + 1/33 + ... + 1/32005)
2A = 1 - 1/32005
A = 1 - 1/32005 / 2
Ủng hộ mk nha ^_-
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow3A-A=2A=1-\frac{1}{3^{2005}}=\frac{3^{2005}-1}{3^{2005}}\)
\(\Rightarrow A=\frac{3^{2005}-1}{2.3^{2005}}\)
đặt B=99/1+99/2+...+1/99
=1+(98/2+1)+(97/3+1)+...+(1/99+1)
=100/100+100/2+...+100/99
=100.(1/2+1/3+...+1/100)
=>A=(1/2+1/3+...+1/100):[100.(1/2+1/3+...+1/100)]
A=1:100=1/100
hok tốt nha
=1/2x2/3x/3/4/.................x2003/2004
= 1x2x3x................x2003
___________________
2x3x4x................x 2004
=1/2004
\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{512}+\frac{1}{1024}=??????????\)
\(< =>1+\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+...+\frac{1}{2\cdot256}+\frac{1}{2\cdot512}\)
\(< =>1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{256}+\frac{1}{2}-\frac{1}{512}\)
\(< =>1+\frac{1}{1}-\frac{1}{512}\)
\(< =>\frac{1023}{512}\)
chuc ban hoc tot nhe :))
đây nhé!!
1+2-3-4+5+6-7-8+9+10-........+2010-2011-2012+2013+2014-2015-2016+2017
=1+(2-3-4+5)+(6-7-8+9)+(10-11-12+13)+....+(2010-2011-2012+2013)+(2014-2015-2016+2017)
=1+0+0+0+.....+0+0
=1.
ĐÚNG THÌ CHO MINK NHA!!^_^
b= 59/10 : 3/2 - [ 21/3 - 41/2 - 2 x 21/3] : 7/4
=9/15 - [ 21/3 - 21/3 x 41/2 - 2]
=9/15 - [ 0 x 41/2 - 2]
=9/15 - 0
=9/15
\(\frac{8}{5}-\frac{1}{2}\div\frac{1}{4}\times\left(\frac{3}{7}+\frac{1}{4}\right)\)
\(=\frac{8}{5}-2\times\frac{19}{28}\)
\(=\frac{8}{5}-\frac{19}{14}\)
\(=\frac{17}{70}\)
\(\frac{8}{5}-\frac{1}{2}:\frac{1}{4}.\left(\frac{3}{7}+\frac{1}{4}\right)=\frac{8}{5}-\frac{1}{2}.4\left(\frac{3}{7}+\frac{1}{4}\right)\)
\(=\frac{8}{5}-2\left(\frac{3}{7}+\frac{1}{4}\right)\)
\(=\frac{8}{5}-\frac{6}{7}+\frac{1}{2}\)
\(=\frac{8.7.2}{5.7.2}-\frac{6.5.2}{5.7.2}+\frac{5.7}{5.7.2}\)
\(=\frac{112}{70}-\frac{60}{70}+\frac{35}{70}\)
\(=\frac{87}{70}\)
k nha Vương Thu Trà
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2003}{2\cdot3\cdot4\cdot...\cdot2004}=\frac{1}{2004}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2003}{2004}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2003}{2\cdot3\cdot4\cdot...\cdot2004}\)
\(A=\frac{1}{2004}\)