Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2015}\right)\)
\(=\left(\frac{2-1}{2}\right)\left(\frac{3-1}{3}\right)\left(\frac{4-1}{4}\right)....\left(\frac{2015-1}{2015}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2013}{2014}.\frac{2014}{2015}\)
\(=\frac{1}{2015}\)
=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2013}{2014}x\frac{2014}{2015}\)
=\(\frac{1x2x3x...x2013x2014}{2x3x4x...x2014x2015}\)
=\(\frac{1}{2015}\)
( Dau x la dau nhan)
Ta có 1+1/2014 +1/x=1/(x+1)+1+1/2013 nên 1/x-1/(x+1)=1/2013-1/(2013+1) nên x=2013
Ta có: \(\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2013}-\frac{2015}{2014}\)
<=> \(\frac{1}{x\left(x+1\right)}=\frac{2014^2-2015.2013}{2013.2014}=\frac{1}{2013.2014}\)
<=> x(x+1)=2013.2014
=> x=2013
Đáp số: x=2013
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
???????????????????????????????????????????????????????????????????????????^_^
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2014}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2013}{2014}\)
\(=\frac{1}{2014}>\frac{1}{2015}\)
=>
\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{2016}{2015}=\frac{3x4x5x...x2016}{2x3x4x....x2015}=\frac{2016}{2}=1008\)
vay C=1008
(1-1/2)x(1-1/3)x.....x(1-1/2014)x(1-1/2015)
=1/2x2/3x.....x2013/2014x2014/2015
=1/2015