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\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) trong đó 3 = x ; 2 = x - 1
\(\frac{1}{\left(x-1\right)x}=\frac{1}{x-1}-\frac{1}{x}\)
ĐẶt A = \(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{30}\cdot\cdot\cdot+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{\left(x-1\right)x}-\frac{2007}{2009}\)
A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}-\frac{2007}{2009}\)
A \(=\frac{1}{2}-\frac{1}{x}-\frac{2007}{2009}\)
a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'
= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009
= 2010/2005
= 402/401
\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)
\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)
\(=\frac{2010}{2005}\)
\(=\frac{402}{401}\)
\(x\)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, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
\(A=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2006}{2007}=\frac{1}{2007}\)
k nha bạn
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)
=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)
=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2010